How do you solve the equation on the interval [0,2pi) for $sinx = cos2x$ ?

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Answer 1 · 2016-06-12T01:02:44

Recall that $cos2theta = 1 - 2sin^2theta$ :

$sinx = 1- 2sin^2x$

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

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

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

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

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

$x = pi/6, (5pi)/6 and (3pi)/2$

Hopefully this helps!

How do you solve the equation on the interval [0,2pi) for sinx = cos2xsinx = cos2x?