Originally Posted by

**prerunner** I have to find all answers to these on the interval 0-2 pi, I think I'm getting them right but I'm not 100% positive

For the first one I had

$\displaystyle cos x-2sinxcosx=0$

and I did

$\displaystyle (2cosx-2sinxcosx)/cosx=0$ Mr F says: Wrong. You have to factorise:

$\displaystyle {\color{red}\cos x (1 - 2 \sin x) = 0 \Rightarrow \cos x = 0}$ or $\displaystyle {\color{red}1 - 2 \sin x = 0}$.

$\displaystyle 1-2sinx=0$

$\displaystyle 1=2sinx$

$\displaystyle sinx=1/2$

and on the interval 0-2pi I got

$\displaystyle pi/6, 5pi/6$ Mr F says: For reasons given above, you're missing solutions.

the second one I have to do

$\displaystyle 2sin^2 x=4sinx+6$

and I got

$\displaystyle sin^2 x=2sinx+3$

$\displaystyle sin^2 x-2sinx-3=0$

$\displaystyle x^2-2x-3=0$ Mr F advises: Use a different symbol to x. x is already being used in the question. Use w = sin x so that w^2 - 2w - 3 = 0.

$\displaystyle (x-3)(x+1)$

$\displaystyle sinx=3 sinx=-1$

and for the interval I got

$\displaystyle 3pi/2$

are there any answers on the interval that I am missing? Mr F says: No.

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