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

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How do you solve $2sin^2x = 2 + cosx$ 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: $2(1-cos^2(x))=2 + cos(x)$ .

Explanation:

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

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

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