How to solve 3cosx+3=sin^2x over the interval (0,2pi]?

Question

$3cosx+3=sin^2x$ Solve over the interval $(0,2pi]$ I got $pi$ as the only solution but can someone check to make sure that I am right? Thanks!

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Answer 1 · 2018-02-13T00:15:20

$x=pi$

.

$3cosx+3=sin^2x$

$3cosx+3=1-cos^2x$

$cos^2x+3cosx+2=0$

Let's factor it:

$(cosx+2)(cosx+1)=0$

$cosx+2=0 :. cosx=-2$ This is not valid because $cos$ values can only be between $1$ and $-1$

$cosx+1=0$

$cosx=-1 :. x=pi$