[SPLIT] solving trig eqns: sin2x(csc2x - 2) = 0, etc

[ilovemath3]

I have a few more questions, and I thank you for any help you may give.

5) sin2x(csc2x - 2) = 0

I have no clue where to start on this one, but I think I would go this route:

2sinxcosx(1 / 2sinxcosx) - 2 = 0

But then they would cancel out leaving me with 2 - 2 = 0 ...?

6) cos (2x - pi/4) = 0

For this one I tried this:

(cos 2x) - (cos pi/4) = 0

I set each of them equal to 0:

cos2x = 0
2x = pi/2 +(2pi)n, so x = (pi/4) + (pi)n
2x = 3pi/2 + (2pi)n, so x = (3pi/4) + (pi)n

I didn't know what to do with the "cos pi/4 = 0".


7) (5sin sq x) + 3sinx = 1

I did:

(5sin sq x) + 3sinx - 1 = 0

I used the Quadratic Formula and that left me with:

(x + 0.238)(x - 0.838) or (x + 0.239)(x + 0.839)

...which would translate into

(sin + 0.238)(sin + 0.839) = 0

But where do I go from there? Becuase if I set them equal to 0, there are no points on the unit circle where the y is equal to 0.238 or 0.839.

any help is GREATLY appreciated, once again thank you for your time and effort

-IloveMath3

 

[tkhunny]

I have no clue where to start on this one, but I think I would go this route:

Do you also have no clue how silly that sounds? I guess you DID have a clue. Why don't you quit thinking like that? You DO have a clue. Get used to it. Your mind is not blank. Figure it out. You're doing just fine -- that means learning!

This sort of problem lends itself to experimentation. You had a good thought and it did not lead you very far. That's fine. Try something else.

Why bother with the double angle formula? What happens when you just solve it?

$\displaystyle sin(2x)\;=\;0\;\Rightarrow\;2x\;=\;0\;+\;k*\pi$ for k an integer

Except that these are NOT solutions. You tell me why.

csc(2x) - 2 = 0
csc(2x) = 2
$\displaystyle sin(2x)\;=\;\frac{1}{2}\;\Rightarrow\;2x\;=\;\pi/6\;+\;2k*\pi\;and\;2x\;=\; 5\pi/6\;+\;2k*\pi$

Notice that if you first multiply through:

sin(2x)*(csc(2x)-2) = 0 ==> 1 - 2*sin(2x) = 0

It is a slightly different problem. Why do you suppose that is?

Hint: Think "Domain".

 

[soroban]

Hello, ilovemath3!

You made some strange moves . . .

I'll assume the answers are between $\displaystyle 0$ and $\displaystyle 2\pi.$

$\displaystyle 5)\;\sin2x(\csc2x\,-\,2) \;= \;0$

It's already factored . . . set each factor equal to zero and solve!

\(\displaystyle \sin2x\:=\:0\;\;\Rightarrow\;\;2x\:=\:0,\,2\pi,\,3\pi,\,4\pi\;\;\Rightarrow\;\;\fbox{x \:=\:0,\,\pi,\,\frac{3\pi}{2},\,2\pi}\)

\(\displaystyle \csc2x\,-\,2\:=\:0\;\;\Rightarrow\;\;\csc2x \:=\:2\;\;\Rightarrow\;\;2x\:=\:\frac{\pi}{6},\,\frac{5\pi}{6},\,\frac{13\pi}{6},\,\frac{17\pi}{6}\;\;\Rightarrow\;\;\fbox{x\:=\:\frac{\pi}{12},\,\frac{5\pi}{12},\,\frac{13\pi}{12},\,\frac{17\pi}{12}}\)

$\displaystyle 6)\;\cos\left(2x\,-\,\frac{\pi}{4}\right) \;= \;0$

For this one I tried this: $\displaystyle \,\cos2x \,-\,\cos\frac{\pi}{4} \:=\:0\;$ . . . illegal!

$\displaystyle \cos\left(2x\,-\,\frac{\pi}{4}\right)\:=\:0\;\;\Rightarrow\;\;2x\,-\,\frac{\pi}{4}\:=\:\frac{\pi}{2},\,\frac{3\pi}{2},\,\frac{5\pi}{2},\,\frac{7\pi}{2}$

. . \(\displaystyle 2x\:=\:\frac{3\pi}{4},\,\frac{7\pi}{4},\,\frac{11\pi}{4},\,\frac{15\pi}{4}\;\;\Rightarrow\;\;\fbox{x\:=\:\frac{3\pi}{8},\,\frac{7\pi}{8},\,\frac{11\pi}{8},\,\frac{15\pi}{8}}\)

$\displaystyle 7)\;5\sin^2x\,+\,3\sin x \:= \:1$

Your use of the Quadratic Formula was correct.
I don't understand your subsequent difficulty.

We have: $\displaystyle \,5(\sin x)^2\,+\,3(\sin x) \,-\,1\;=\;0$

Quadratic Formula: $\displaystyle \sin x \;= \;\frac{-3\,\pm\,\sqrt{3^2\,-\,4(5)(-1)}}{2(5}\;=\;\frac{-3\,\pm\,\sqrt{29}}{10} \;\approx\;\{0.2385,\:-0.8385\}$

\(\displaystyle \sin x \:=\:0.2385\;\;\Rightarrow\;\;x\:=\:\sin^{-1}(0.2385)\;\;\Rightarrow\;\;\fbox{x\:=\:13.8^o,\:166.2^o}\)

\(\displaystyle \sin x \:=\:-0.8385\;\;\Rightarrow\;\;x\:=\:-57^o\;\;\Rightarrow\;\;\fbox{x\:=\:237^o,\:303^o}\)