How do you solve $3 cos x + 3 = 2 sin ^2 x$ over the interval 0 to 2pi?

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Answer 1 · 2016-02-12T07:31:01

$3 cos x + 3 = 2 sin ^2 x$

We know that

$Sin^2 x + cos ^ 2 x= 1$
$=>sin^2 x = 1 - cos^2 x$

Lets substitute this in our equation

$3 cos x + 3 = 2(1 - cos^2x)$

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

Bring everything over to one side

$2cos^2 x + 3 cos x + 1 = 0$

$Let cos x = a$

$=> 2a^2 + 3a + 1 = 0$

$alpha + beta = -3/2$
$alpha*beta = 1/2$

Solving we get

The roots are

$alpha = -1$

$beta = -1/2$

$therefore cos x = -1 or -1/2$

$=>x = 180 or 120$
$=>x = pi or 2/3 pi$

These both lie in the given interval and hence are correct solutions

How do you solve 3 cos x + 3 = 2 sin ^2 x3 cos x + 3 = 2 sin ^2 x over the interval 0 to 2pi?