Need help with 6 Trig-solving problems

[frosthawk97]

Hello! Fellow Precalculus student trying to complete some problems for test corrections. There are six problems that I am struggling with and it would be greatly appreciated if I could get some help.

1) Solve 4sin(3x) = (-1/2) over the reals.

So far, I have worked down the problem to:
sin(3x) = (-1/8) + 2kπ (divide by 4)
(I'm assuming I am looking for general solutions as I'm solving "over the reals," but please correct me if I'm suppose to be finding specific solutions)
3x = (-1/8) + 2kπ
x= (-1/24) + ((2kπ)/3)

For this question, I am wondering is my answer correct at all? Did I do everything correctly, or is there something I've missed? I can't find where sin(x) = -1/8 anywhere on the unit circle, obviously, but I'm not sure if I find the exact value and add ((2kπ)/3) to it or not.

2) What is the set of solutions for sin(2x) = 3sin(x) over the interval of [0, 2π)?

So far, I have worked it down to:
sin(2x) = 2sin(x)cos(x); so the equation is now 2sin(x)cos(x) = 3sin(x)

2sin(x)cos(x) - 3sin(x) = 0

Factor to: sin(x) * (2cos(x) - 3) = 0

Two solutions: sin(x) = 0 & (2cos(x) -3) = 0

sin(x) = 0 at 0 and π on the unit circle (and I'm assuming these are two solutions I have)

(2cos(x)-3) = 0
2cos(x)= 3
cos(x)= (-3/2)
no solution
*I'm assuming there are no solutions due to the fact cosine never equals (-3/2)--since it's range is [-1,1]

So is the answer 2 solutions?

3) How many solutions does sin(x/2) = +/- 1 (positive and negative 1) over the interval [0, 2π)?

Because it's asking for both positive and negative 1, I know that sin(x) equals positive 1 at (π/2) and negative 1 at (3π/2). But if I'm not mistaken, isn't there a domain restriction for sine at [(-π/2), (π/2)]? Furthermore, does this domain restriction even apply? I know I'll have to multiply whatever solution I get by 2, and I know (π/2) * 2 = π.

4) What is the set of all solutions for cos(x)tan(x)sin(x) = sin²x?

So far I've worked this out to:

cos(x)tan(x)sin(x) = (cos(x)sin²(x)) /cos(x)

(cos(x)sin²(x))/cos(x) = sin²(x)

sin²(x) = cos²(x)-1

(cos(x)cos²(x)-1)/cos(x) = cos²(x)-1

And this is where I'm stuck at. I've tried subtracting cos²​(x)-1 from the right side and figured I'd factor everything on the left but that hasn't worked out. I thought I could do an alternate route where I just converted the equation on the left to sine and cosine and then subtracted sin²(x) on both sides and give it a common denominator and factor out the equation to solve, but that lead me to no avail as well.

5) How many solutions does tan(5x) = √3 have over [0, 2π)?
For this one, I know that tan(x) = √3 at (π/3) and (4π/3). So, in order to find the general solution formula for each I'd have to set it up where
tan(5x) = (π/3) + kπ and tan(5x) = (4π/3) + kπ.

5x = (π/3) + kπ
x= (π/15) + ((kπ)/5)

5x = (4π/3) + kπ
x= (4π/15) + ((kπ)/5)

So I'm guessing at each interval I just add 3π/15 to each interval until you reach the end of the range?
(π/15), (4π/15), (7π/15), (2π/3), (13π/15), (16π/15), (18π/15), (19π/15), (21π/15), (22π/15), (24π/15), (5π/3), (27π/15), (28π/15), (2π)

But something doesn't feel right. I feel like I don't need to count all of these solutions and can simplify to one statement, but I'm not quite sure how--if that is the case.

​6) Find all real solutions for x when sin(x)cos(π) - cos(x)sin(π) = (-√3/2).

I know that this is a cosine double angle identity, so I can rewrite the equation as:

cos(x + π) = (-√3/2)

I know cos(x) = (-√3/2) at (π/6) and (11π/6). So:

cos(x) = ((π/6) - π ) + 2kπ

x = (π/6) - (6π/6) +2kπ

x = (-5π/6) + 2kπ

cos(x) = ((11π/6) - π) + 2kπ

x = (5π/6) + 2kπ

And do I just leave things as is since I'm assuming "all real solutions" means that I'm looking for a general solution?


-Thanks in advance to anyone who helps out~

 

[Harry_the_cat]

Q1

1) Solve 4sin(3x) = (-1/2) over the reals.

So far, I have worked down the problem to:
sin(3x) = (-1/8) + 2kπ (divide by 4)

Don't add 2k(pi) at this stage. 2k(pi) is added to account for coterminal solutions. You haven't found any solutions yet.

Best way is to let 3x = A,
and solve
4 sin(A) = -1/2
sin(A) = -1/8 (dividing by 4 was correct)

Now arcsin (1/8) = 0.1253 (ignoring sign for now)
Since sin (A) is neg, A must be in 3rd or 4th quadrant

so

A = (pi) + 0.1253 + 2k(pi) or A = 2(pi) - 0.1253 + 2k(pi) ... ("over the reals" means general solutions here and also answers in radians)

So now you've solved for A.

Now substitute 3x for A and continue on to find x.


Q2. Yes that's correct. Just two solutions in [ 0, 2pi ]....x = 0 or pi. cos x = 3/2 yields no solutions. (You have the right reason but note that cos x is 3/2 not -3/2.


Q3. The domain restriction that applies here is [0, 2pi) as stated in the question.
Again, let x/2 = A and solve first for A.

Note that if 0 <= x < 2pi, then 0<= x/2 < pi ie 0<= A < pi
So, sin A = +/- 1, yields A= pi/2 only since 0<= A < pi
Now you've solved for A, replace A = x/2
ie x/2 = pi/2 so x = pi


Q4

So far I've worked this out to:

cos(x)tan(x)sin(x) = (cos(x)sin²(x)) /cos(x)

(cos(x)sin²(x))/cos(x) = sin²(x)

Ok so it seems that cos(x)tan(x)sin(x) = sin²(x) all the time BUT you can only cancel a common factor if it doesn't equal 0.
Your second statement is true only if cos x =/= 0 ... ie doesn't equal 0

So the equation cos(x)tan(x)sin(x) = sin²(x) is true for all values of x except when cos x = 0 (which happens when x = ? +2k(pi) or x = ? +2k(pi))

 

[Deleted member 4993]

4) What is the set of all solutions for cos(x)tan(x)sin(x) = sin²x?

I would start it the following way:

cos(x)*tan(x)*sin(x) - sin²(x) = 0

sin(x) * [cos(x)*tan(x) - sin(x)] = 0

sin(x) = 0 → x = n * π .... (n = 0, ±1, ±2, ±3....) ................................................................................ (1)

cos(x)*tan(x) - sin(x) = 0 for all values of 'x' except x = (2n+1)/2 * π .... (n = 0, ±1, ±2, ±3....) ................. (2)

answer set (1) is a subset of answer set (2). So .............

 

[Harry_the_cat]

5) How many solutions does tan(5x) = √3 have over [0, 2π)?
For this one, I know that tan(x) = √3 at (π/3) and (4π/3). Correct
So, in order to find the general solution formula for each I'd have to set it up where
tan(5x) = (π/3) + kπ and tan(5x) = (4π/3) + kπ. (You should no longer have the tan in this line.)

I think you mean 5x = (π/3) + kπ and tan(5x) = (4π/3) + kπ. You are doubling up here!

What you should have is 5x = (π/3) + 2kπ and tan(5x) = (4π/3) + 2kπ to account for coterminal angles.

Draw a diagram and label where π/3 and 4π/3 are. Notice the angle between them is π.

So you could write these two statements as 5x = (π/3) + kπ . If k is even, you will get angles coterminal with π/3, if k is odd you will get angles coterminal with 4π/3.

So,
5x = (π/3) + kπ
x= (π/15) + ((kπ)/5)

So let k=0, 1, 2 etc to find all the solutions in[0,2π). Do this and you will see you are wrong to just add a constant interval. There will be 10 solutions.

Note also, the question didn't ask you to find the all, just to say how many there were.

It makes sense (think about it) that if tan(x) = √3 has 2 solutions in [0, 2π), then tan(2x)= √3 will have 4 solutions,....tan (5x) = √3 will have 10.

6) Find all real solutions for x when sin(x)cos(π) - cos(x)sin(π) = (-√3/2).

I know that this is a cosine double angle identity, so I can rewrite the equation as:

cos(x + π) = (-√3/2) No! Wrong rule! Try again from here.

I know cos(x) = (-√3/2) at (π/6) and (11π/6). So: (but cos x is positive in these quadrants!)

cos(x) = ((π/6) - π ) + 2kπ

x = (π/6) - (6π/6) +2kπ

x = (-5π/6) + 2kπ

cos(x) = ((11π/6) - π) + 2kπ

x = (5π/6) + 2kπ

And do I just leave things as is since I'm assuming "all real solutions" means that I'm looking for a general solution Yes

 

[Prove It]

Hello! Fellow Precalculus student trying to complete some problems for test corrections. There are six problems that I am struggling with and it would be greatly appreciated if I could get some help.

1) Solve 4sin(3x) = (-1/2) over the reals.

So far, I have worked down the problem to:
sin(3x) = (-1/8) + 2kπ (divide by 4)
(I'm assuming I am looking for general solutions as I'm solving "over the reals," but please correct me if I'm suppose to be finding specific solutions)
3x = (-1/8) + 2kπ
x= (-1/24) + ((2kπ)/3)

For this question, I am wondering is my answer correct at all? Did I do everything correctly, or is there something I've missed? I can't find where sin(x) = -1/8 anywhere on the unit circle, obviously, but I'm not sure if I find the exact value and add ((2kπ)/3) to it or not.

2) What is the set of solutions for sin(2x) = 3sin(x) over the interval of [0, 2π)?

So far, I have worked it down to:
sin(2x) = 2sin(x)cos(x); so the equation is now 2sin(x)cos(x) = 3sin(x)

2sin(x)cos(x) - 3sin(x) = 0

Factor to: sin(x) * (2cos(x) - 3) = 0

Two solutions: sin(x) = 0 & (2cos(x) -3) = 0

sin(x) = 0 at 0 and π on the unit circle (and I'm assuming these are two solutions I have)

(2cos(x)-3) = 0
2cos(x)= 3
cos(x)= (-3/2)
no solution
*I'm assuming there are no solutions due to the fact cosine never equals (-3/2)--since it's range is [-1,1]

So is the answer 2 solutions?

3) How many solutions does sin(x/2) = +/- 1 (positive and negative 1) over the interval [0, 2π)?

Because it's asking for both positive and negative 1, I know that sin(x) equals positive 1 at (π/2) and negative 1 at (3π/2). But if I'm not mistaken, isn't there a domain restriction for sine at [(-π/2), (π/2)]? Furthermore, does this domain restriction even apply? I know I'll have to multiply whatever solution I get by 2, and I know (π/2) * 2 = π.

4) What is the set of all solutions for cos(x)tan(x)sin(x) = sin²x?

So far I've worked this out to:

cos(x)tan(x)sin(x) = (cos(x)sin²(x)) /cos(x)

(cos(x)sin²(x))/cos(x) = sin²(x)

sin²(x) = cos²(x)-1

(cos(x)cos²(x)-1)/cos(x) = cos²(x)-1

And this is where I'm stuck at. I've tried subtracting cos²​(x)-1 from the right side and figured I'd factor everything on the left but that hasn't worked out. I thought I could do an alternate route where I just converted the equation on the left to sine and cosine and then subtracted sin²(x) on both sides and give it a common denominator and factor out the equation to solve, but that lead me to no avail as well.

5) How many solutions does tan(5x) = √3 have over [0, 2π)?
For this one, I know that tan(x) = √3 at (π/3) and (4π/3). So, in order to find the general solution formula for each I'd have to set it up where
tan(5x) = (π/3) + kπ and tan(5x) = (4π/3) + kπ.

5x = (π/3) + kπ
x= (π/15) + ((kπ)/5)

5x = (4π/3) + kπ
x= (4π/15) + ((kπ)/5)

So I'm guessing at each interval I just add 3π/15 to each interval until you reach the end of the range?
(π/15), (4π/15), (7π/15), (2π/3), (13π/15), (16π/15), (18π/15), (19π/15), (21π/15), (22π/15), (24π/15), (5π/3), (27π/15), (28π/15), (2π)

But something doesn't feel right. I feel like I don't need to count all of these solutions and can simplify to one statement, but I'm not quite sure how--if that is the case.

​6) Find all real solutions for x when sin(x)cos(π) - cos(x)sin(π) = (-√3/2).

I know that this is a cosine double angle identity, so I can rewrite the equation as:

cos(x + π) = (-√3/2)

I know cos(x) = (-√3/2) at (π/6) and (11π/6). So:

cos(x) = ((π/6) - π ) + 2kπ

x = (π/6) - (6π/6) +2kπ

x = (-5π/6) + 2kπ

cos(x) = ((11π/6) - π) + 2kπ

x = (5π/6) + 2kπ

And do I just leave things as is since I'm assuming "all real solutions" means that I'm looking for a general solution?


-Thanks in advance to anyone who helps out~

This is the very reason why you shouldn't ever post any more than two questions per thread, as it becomes very convoluted and difficult to follow!

It appears that you have received help on each question, so I'll put a summary of each question in this thread.

Q.1
\(\displaystyle \displaystyle \begin{align*} 4\sin{ \left( 3\,x \right) } &= -\frac{1}{2} \\ \sin{ \left( 3\,x \right) } &= -\frac{1}{8} \\ 3\,x &= \left\{ \pi + \arcsin{ \left( \frac{1}{8} \right) } , 2\,\pi - \arcsin{ \left( \frac{1}{8} \right) } \right\} + 2\,\pi\,n \textrm{ where } n \in \mathbf{Z} \\ x &= \left\{ \frac{\pi}{3} + \frac{1}{3}\arcsin{ \left( \frac{1}{8} \right) } , \frac{2\,\pi}{3} - \frac{1}{3} \arcsin{ \left( \frac{1}{8} \right) } \right\} + \frac{2\,\pi\,n}{3} \end{align*}\)

Q.2
\(\displaystyle \displaystyle \begin{align*} \sin{ \left( 2\,x \right) } &= 3\sin{(x)} \textrm{ where } x \in [ 0, 2\,\pi ) \\ 2\sin{(x)}\cos{(x)} &= 3\sin{(x)} \\ 2\sin{(x)}\cos{(x)} - 3\sin{(x)} &= 0 \\ \sin{(x)} \left[ 2\cos{(x)} - 3 \right] &= 0 \\ \sin{(x)} &= 0 \textrm{ or } 2\cos{(x)} - 3 = 0 \\ x &= n\,\pi \textrm{ where } n \in \mathbf{Z} \textrm{ (notice that since } -1 \leq \cos{(x)} \leq 1 \textrm{ for all } x \textrm{ that means there are no solutions to the second of these equations...)} \\ x &= \left\{ 0, \pi \right\} \textrm{ in the domain } x \in [0, 2\,\pi ) \end{align*}\)

Q.3
\(\displaystyle \displaystyle \begin{align*} \sin{ \left( \frac{x}{2} \right) } &= \pm 1 \textrm{ where } x \in [0, 2\,\pi ) \\ \frac{x}{2} &= \frac{\left( 2\,n + 1 \right) \, \pi}{2} \textrm{ where } n \in \mathbf{Z} \\ x &= \left( 2\,n + 1 \right) \,\pi \\ x &= \pi \textrm{ when } x \in [0, 2\,\pi ) \end{align*}\)

So there is only one solution.

Q.4
Solve \(\displaystyle \displaystyle \begin{align*} \cos{(x)}\tan{(x)}\sin{(x)} &= \sin^2{(x)} \end{align*}\)

Notice that

\(\displaystyle \displaystyle \begin{align*} \cos{(x)}\tan{(x)}\sin{(x)} &\equiv \cos{(x)}\,\frac{\sin{(x)}}{\cos{(x)}}\,\sin{(x)} \textrm{ where } x \in \mathbf{R} \backslash \left\{ \frac{\left( 2\,n + 1 \right) \, \pi}{2} \right\} \textrm{ where } n \in \mathbf{Z} \\ &\equiv \sin^2{(x)} \end{align*}\)

So the equation is true no matter what x goes in (provided it is in the domain), thus the solution is \(\displaystyle \displaystyle \begin{align*} x \in \mathbf{R} \backslash \left\{ \frac{\left( 2\,n + 1 \right) \,\pi}{2} \right\} \textrm{ where } n \in \mathbf{Z} \end{align*}\)

Q.5
\(\displaystyle \displaystyle \begin{align*} \tan{ \left( 5\,x \right) } &= \sqrt{3} \textrm{ where } x \in \left[ 0 , 2\,\pi \right) \\ x &= \frac{\pi}{15} + \frac{ \pi \, n }{5} \\ x &= \frac{ \left( 1 + 3\,n \right) \,\pi }{15} \\ x &= \left\{ \frac{ \pi }{15} , \frac{ 4 \, \pi }{15} , \frac{ 7 \, \pi }{15} , \frac{ 2 \, \pi }{3} , \frac{ 13 \, \pi }{15} , \frac{ 16 \, \pi }{15} , \frac{ 19 \, \pi }{15} , \frac{ 22 \, \pi }{15} , \frac{ 5 \, \pi }{3} , \frac{ 28 \, \pi }{15} \right\} \textrm{ when } x \in [ 0 , 2 \, \pi )
\end{align*}\)

So 10 solutions in that region.

Q.6
\(\displaystyle \displaystyle \begin{align*} \sin{(x)}\cos{ \left( \pi \right) } - \cos{(x)}\sin{ \left( \pi \right) } &= -\frac{\sqrt{3}}{2} \end{align*}\)

Notice that \(\displaystyle \displaystyle \begin{align*} \sin{ \left( \alpha + \beta \right) } \equiv \sin{ \left( \alpha \right) } \cos{ \left( \beta \right) } - \cos{ \left( \alpha \right) } \sin{ \left( \beta \right) } \end{align*}\) so that means \(\displaystyle \displaystyle \begin{align*} \sin{(x)} \cos{ \left( \pi \right) } - \cos{ (x) } \sin{ \left( \pi \right) } \equiv \sin{ \left( x + \pi \right) } \end{align*}\), giving

\(\displaystyle \displaystyle \begin{align*} \sin{ \left( x + \pi \right) } &= -\frac{\sqrt{3}}{2} \end{align*}\)

and through symmetry we have \(\displaystyle \displaystyle \begin{align*} \sin{ \left( x + \pi \right) } \equiv -\sin{(x)} \end{align*}\) so that means

\(\displaystyle \displaystyle \begin{align*} -\sin{(x)} &= -\frac{\sqrt{3}}{2} \\ \sin{(x)} &= \frac{\sqrt{3}}{2} \\ x &= \left\{ \frac{\pi}{3} , \pi - \frac{\pi}{3} \right\} + 2\,\pi\,n \textrm{ where } n \in \mathbf{Z} \\ x &= \left\{ \frac{\pi}{3}, \frac{2\,\pi}{3} \right\} + 2\,\pi\,n \end{align*}\)

 

[frosthawk97]

This is the very reason why you shouldn't ever post any more than two questions per thread, as it becomes very convoluted and difficult to follow!

Oh no, I'm so sorry! I won't do it again next time, I promise.

But regardless, thank you all for the help!