Trig. equations of type sin A = sin B

Qwertyuiop[]

Qwertyuiop[]

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I have come across these 2 questions that i don't know how to solve. [imath]sin\left(x+\frac{\pi }{3}\right)=sin\left(2x\right)[/imath] & [imath]sin (x + \pi/6) = sin (2x + 3\pi/4)[/imath] . When I usually get something like this where there is something inside the brackets , I just set the bracket function equal to let's say, theta and solve for theta and then substitute to find the values of x. So If I have sin (2x + pi/3) = 1/2 I will write this as sin (theta) = 1/2 where theta = (2x + pi/3). Now the problem here is that we have like x + pi/3 and 2x and for the 2nd question x+ pi/6 & 2x + 3pi/4 . What do I do here, can't do 2 substitutions I will end up with 2 variables. I also tried writing sin 2x as 2 sinx cosx but then I get a more complicated expression and then don't know how to proceed, tried factoring but no luck.
So what's the correct approach to this question ?
 
Do you know how to convert a difference of two trigonometric functions into a product of other trigonometric functions?
 
Qwertyuiop[] said:
I have come across these 2 questions that i don't know how to solve. [imath]sin\left(x+\frac{\pi }{3}\right)=sin\left(2x\right)[/imath] & [imath]sin (x + \pi/6) = sin (2x + 3\pi/4)[/imath] . When I usually get something like this where there is something inside the brackets , I just set the bracket function equal to let's say, theta and solve for theta and then substitute to find the values of x. So If I have sin (2x + pi/3) = 1/2 I will write this as sin (theta) = 1/2 where theta = (2x + pi/3). Now the problem here is that we have like x + pi/3 and 2x and for the 2nd question x+ pi/6 & 2x + 3pi/4 . What do I do here, can't do 2 substitutions I will end up with 2 variables. I also tried writing sin 2x as 2 sinx cosx but then I get a more complicated expression and then don't know how to proceed, tried factoring but no luck.
So what's the correct approach to this question ?
Click to expand...
You are trying to paraphrase your original problem/s.

Are these two separate problems or parts of one problem?
 
Qwertyuiop[] said:
I have come across these 2 questions that i don't know how to solve. [imath]sin\left(x+\frac{\pi }{3}\right)=sin\left(2x\right)[/imath] & [imath]sin (x + \pi/6) = sin (2x + 3\pi/4)[/imath] . When I usually get something like this where there is something inside the brackets , I just set the bracket function equal to let's say, theta and solve for theta and then substitute to find the values of x. So If I have sin (2x + pi/3) = 1/2 I will write this as sin (theta) = 1/2 where theta = (2x + pi/3). Now the problem here is that we have like x + pi/3 and 2x and for the 2nd question x+ pi/6 & 2x + 3pi/4 . What do I do here, can't do 2 substitutions I will end up with 2 variables. I also tried writing sin 2x as 2 sinx cosx but then I get a more complicated expression and then don't know how to proceed, tried factoring but no luck.
So what's the correct approach to this question ?
Click to expand...
.problem(1)....... [imath]sin\left(x+\frac{\pi }{3}\right)=sin\left(2x\right)[/imath]........................... problem(2)....... [imath]sin (x + \pi/6) = sin (2x + 3\pi/4)[/imath]

x + π/3 = 2*x +2*K * π .............or ...........x + π/3 = (2n+1)*π - 2*x .........similarly...........x + π/3 = 2*x +2*K * π ..........or ........x + π/3 = (2n+1)*π - 2*x

For two problems you get different sets of solutions. Can you please explain your problem a little more!!!
 
Subhotosh Khan said:
.problem(1)....... [imath]sin\left(x+\frac{\pi }{3}\right)=sin\left(2x\right)[/imath]........................... problem(2)....... [imath]sin (x + \pi/6) = sin (2x + 3\pi/4)[/imath]

x + π/3 = 2*x +2*K * π .............or ...........x + π/3 = (2n+1)*π - 2*x ............similarly.............x + π/3 = 2*x +2*K * π .............or ...........x + π/3 = (2n+1)*π - 2*x

For two problems you get different sets of solutions. Can you please explain your problem a little more!!!
Click to expand...
In an example, they use sum to product formula to solve a similar problem. This method might be too advanced and they are not using this to solve the question so I think I don't need to know this for now.
 
Dr.Peterson said:
Try here:

en.wikipedia.org

List of trigonometric identities - Wikipedia

en.wikipedia.org en.wikipedia.org

SK's way is more direct (and it's what I saw first), but this may make it a little easier, and is worth knowing about.
Click to expand...
They use this product to sum formula : sin p - sin q = 2 cos ((p + q) /2) sin ((p-q) /2). If I apply this to the first question [imath]sin (x + \pi/3) = sin(2x)[/imath] this simplifies to [math]2 cos ( (3x + \pi/3)/ 2) sin ( (-x + \pi/3)/ 2)[/math] now where do I go from here ?
 
You forgot to turn this expression into an equation, i.e., this whole thing must be equal to zero.
 
blamocur said:
You forgot to turn this expression into an equation, i.e., this whole thing must be equal to zero.
Click to expand...
2cos((3x+π/3)/2)sin((−x+π/3)/2) = 0 and then equate them both to 0?
2cos((3x+π/3)/2) = 0 or sin((−x+π/3)/2) = 0 ===> cos((3x+π/3)/2) = 0 or sin((−x+π/3)/2) = 0 ?
 
Qwertyuiop[] said:
In an example, they use sum to product formula to solve a similar problem. This method might be too advanced and they are not using this to solve the question so I think I don't need to know this for now.
Click to expand...
I wouldn't say too advanced; but it is not necessary to use it.
Qwertyuiop[] said:
2cos((3x+π/3)/2)sin((−x+π/3)/2) = 0 and then equate them both to 0?
2cos((3x+π/3)/2) = 0 or sin((−x+π/3)/2) = 0 ===> cos((3x+π/3)/2) = 0 or sin((−x+π/3)/2) = 0 ?
Click to expand...
Yes. Just do it!

Then try SK's method and see if (a) they give the same result, and (b) one is easier than the other. That's a good way to learn.
 
Dr.Peterson said:
I wouldn't say too advanced; but it is not necessary to use it.

Yes. Just do it!

Then try SK's method and see if (a) they give the same result, and (b) one is easier than the other. That's a good way to learn.
Click to expand...
I tried the with the sum to product formula for both questions , I get the correct answers ! Verified with Desmos . But for the 2nd question , I have one missing solution.
SK method I got stuck before even starting :LOL: . The sum to product is easier for me so i will stick to this one.
 
Making a general point

[math]\text {Find all values of } x \text { such that } f(g(x)) = f(h(x)).\\ \text {By the DEFINITION of a function, if there exists } a \text { such that } g(a) = h(a),\\ \text {then } f(g(a)) = f(h(a)).[/math]
Do you follow that?

So one way to look for solutions is to find the zeroes of [imath]i(x) = g(x) - h(x).[/imath]

There are two problems with this method. The first is that i(x) may have no zeroes. The second is that there may be additional solutions over and above those zeroes. SK combined some steps. On problem 1, can you solve

[math]x + \dfrac{\pi}{3} - 2x = 0.[/math]
That gives you ONE solution. Are there others? Remember f(x) in this problem is a sine function, which is periodic. What does that tell you?
 
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You need to know that sinA = sinB means A = B + 2pi*k or A-B = 2pi*k. Any other solutions to sinA = sinB??

sin(x+π/3)=sin(2x)

(2x) - (x+π/3i) = x-π/3 = 2pi*k
x = π/3 + 2pi*k
 
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