How do you solve $4sin^2 x - 3cosx -3 = 0$ ?

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Answer 1 · 2015-04-23T17:41:55

We would rather have just one trigonometric function if possible.
How are $sinx$ and $cosx$ (the two functions we see here) related?

Yes, $sin^2x+cos^2x=1$ , so we can replace $sin^2x$ with $1-cos^2x$ .

$4sin^2 x - 3cosx -3 = 0$

$4(1-cos^2 x) - 3cosx -3 = 0$

So we need to solve:

$-4cos^2x-3cosx+1=0$ , or

$4cos^2x+3cosx-1=0$

Solve for $cosx$ first,
then for $x$

How do you solve 4sin^2 x - 3cosx -3 = 04sin^2 x - 3cosx -3 = 0?