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

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

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Answer 1 · 2016-03-28T22:02:12

First, perform a Pythagorean substitution to remove the sine term from the left side: $2 (1 - {cos}^{2} (x)) = 2 + cos (x)$ $2(1-cos^2(x))=2 + cos(x)$ .

Explanation:

Simplify the left side : $2 - 2 {cos}^{2} (x) = 2 + cos (x)$ $2-2cos^2(x)=2+cos(x)$
Gather like terms and set equal to 0: $0 = 2 {cos}^{2} (x) + cos (x)$ $0=2cos^2(x)+cos(x)$
Factor the right side: $0 = cos (x) (2 cos (x) + 1)$ $0=cos(x)(2cos(x) + 1)$
Use the Zero Product Property:
$cos (x) = 0$ $cos(x) = 0$ or $2 cos (x) + 1 = 0$ $2cos(x)+1=0$
$cos (x) = 0$ $cos(x)=0$ or $2 cos (x) = - 1$ $2cos(x) = -1$
$cos (x) = 0$ $cos(x)=0$ or $cos (x) = - \frac{1}{2}$ $cos(x) = -1/2$
So, $x = {cos}^{- 1} (0)$ $x = cos^-1(0)$ or $x = {cos}^{- 1} (- \frac{1}{2})$ $x = cos^-1(-1/2)$
x = $\frac{π}{2}$ $pi/2$ , $\frac{3 \cdot π}{2}$ $(3*pi)/2$ , or $\frac{2 \cdot π}{3}$ $(2*pi)/3$ , $\frac{4 \cdot π}{3}$ $(4*pi)/3$

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

1 Answer

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