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

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Answer 1 · 2016-02-12T07:31:01

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

We know that

${sin}^{2} x + {cos}^{2} x = 1$ $Sin^2 x + cos ^ 2 x= 1$
$\Rightarrow {sin}^{2} x = 1 - {cos}^{2} x$ $=>sin^2 x = 1 - cos^2 x$

Lets substitute this in our equation

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

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

Bring everything over to one side

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

$L e t cos x = a$ $Let cos x = a$

$\Rightarrow 2 a^{2} + 3 a + 1 = 0$ $=> 2a^2 + 3a + 1 = 0$

$α + β = - \frac{3}{2}$ $alpha + beta = -3/2$
$α \cdot β = \frac{1}{2}$ $alpha*beta = 1/2$

Solving we get

The roots are

$α = - 1$ $alpha = -1$

$β = - \frac{1}{2}$ $beta = -1/2$

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

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

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

How do you solve 3 cos x + 3 = 2 sin ^2 x3cosx+3=2sin2x3 cos x + 3 = 2 sin ^2 x over the interval 0 to 2pi?