I'm wrong. My solution is wrong I think. Please help me. Question below

Question

owlet · Answer

My solution is also iincluded

owlet · Answer

question: Solve $\sf csc^2 x -1 = 4 cos^2x$

solomonzelman · Answer

$\large\color{black}{ \displaystyle \csc^2x-1=4\cos^2x  }$

$\large\color{black}{ \displaystyle \cot^2x=4\cos^2x  }$

$\large\color{black}{ \displaystyle \left[\cos^2x~/~\sin^2x\right]=4\cos^2x  }$

$\large\color{black}{ \displaystyle 1/\sin^2x=4  }$

$\large\color{black}{ \displaystyle 1/\sin^2x=1/{~~}(^1/_4)  }$

$\large\color{black}{ \displaystyle \sin^2x=1/4  }$

solomonzelman · Answer

this is what I would start off.

owlet · Answer

then take the square root right? so you'll also end up with sin x = positive/negative 1/2

owlet · Answer

oh i think i get it.. I'm missing sin x= 1/2 which is Pi/6 

so my solution is really messy. I'll try to do it again. thanks!

owlet · Answer

and 5pi/6 i think

solomonzelman · Answer

* sin(x) = ½ *         or         * sin(x) = - ½ *

parthkohli · Answer

$$\cos (x) = 0$$

triciaal · Answer

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