Solution for (cos2x)(csc^2x)=2cos2x & Sinx=cos2x?

[Timcago]

Find all solutions of the following equations, in the given intervals.
A) (cos2x)(csc^2x)=2cos2x, x = [0,2)
B) Sinx=cos2x, all real solutions

What do they want me to do here?

 

[soroban]

Hello, Timcago!

Find all solutions of the following equations, in the given intervals.

$\displaystyle A)\;(\cos2x)(\csc^2x)\:=\:2\cdot\cos2x,\;\;x = [0,2)$

$\displaystyle B)\;\sin x\:=\:\cos2x,\;\;$all real solutions

What do they want me to do here? . . . How about "solve for x"?

A) \(\displaystyle \(\cos2x)(\csc^2x)\:=\:2\cdot\cos2x\;\) on the interval [0,2)

Then: \(\displaystyle \\cos2x)(\csc^2x)\,-\,2\cdot\cos2x\;=\;0\)

Factor: $\displaystyle \:\cos2x\left(\csc^2x\,-\,2\right)\;=\;0$


And we have two equations to solve:

$\displaystyle \;\;\cos2x\:=\:0\;\;\Rightarrow\;\;2x\,=\,\frac{\pi}{2},\;\frac{3\pi}{2}\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{4},\;\frac{3\pi}{4}$

$\displaystyle \;\;\csc^2x\,=\,2\;\;\Rightarrow\;\;\csc x\,=\,\pm\sqrt{2}\;\;\Rightarrow\;\;\sin x\,=\,\pm\frac{1}{\sqrt{2}}\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{4},\;\frac{3\pi}{4}$

But $\displaystyle \frac{3\pi}{4}\,>\,2$ . . . Therefore, the only solution is: $\displaystyle \,x\,=\,\frac{\pi}{4}$


B) $\displaystyle \sin x \:=\:\cos2x\;$ ... all real solutions

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

Factor: $\displaystyle \,(2\sin x\,-\,1)(\sin x\,+\,1)\:=\:0$


And we have two equations to solve:

$\displaystyle \;\;2\sin x\,-\,1\:=\:0\;\;\Rightarrow\;\;\sin x\,=\,\frac{1}{2}\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{6}\,+\,2\pi n,\;\frac{5\pi}{6}\,+\,2\pi n$

$\displaystyle \;\;\sin x\,+\,1\:=\:0\;\;\Rightarrow\;\;\sin x\,=\,-1\;\;\Rightarrow\;\;x\,=\,\frac{3\pi}{2}\,+\,2\pi n$

 

[Timcago]

ohhhhh yea, i remember doing those now. I am doing the review for the final exam right now and i completely blanked when i saw that.

I had a typo. The interval is actually [0,2*PI)

Anyways on part A we have sinx=Plus or minus the root of 2 over 2.

Wouldnt that include PI/4, 3PI/4, 5PI/4, 7PI/4 since its both positive and negative root of 2 over 2?