How do you solve $cos2x+sinx=0$ and find all solutions in the interval $[0,2pi)$ ?

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Answer 1 · 2016-10-27T20:22:24

Use the identity $cos2x = 1- 2sin^2x$ to get rid of all terms in cosine.

$1 - 2sin^2x + sinx = 0$

$-2sin^2x + sinx + 1 = 0$

$-2sin^2x + 2sinx - sinx + 1 = 0$

$-2sinx(sinx - 1) - (sinx - 1) = 0$

$(-2sinx - 1)(sinx - 1) = 0$

$sinx = -1/2 and sinx =1$

$x = (7pi)/6, (11pi)/6, pi/2$

Hopefully this helps!

How do you solve cos2x+sinx=0cos2x+sinx=0 and find all solutions in the interval [0,2pi)[0,2pi)?