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

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

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