How do you solve $2sin^2x-cosx=1$ on the interval [0,2pi]?

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Answer 1 · 2016-02-19T13:35:21

${pi/3, pi, {5pi}/3}$

Use the identity

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

to make the equation into a quadratic equation w.r.t. $cosx$ . Then proceed to solve the quadratic by factorization/completing the square.

$2sin^2x + cosx = 2(1 - cos^2x) - cosx$

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

$= 1$

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

$(2cosx - 1)(cosx + 1) = 0$

This means that

$cosx = 1/2$ $or$ $cosx = -1$

$cosx$ is positive for $0 < x < pi/2$ $and$ ${3pi}/2 < x < 2pi$ .

$x = pi/3 or pi or {5pi}/3$

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