How do you solve $sin2x-cosx=0$ ?

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Answer 1 · 2016-11-25T18:52:04

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

Before we solve, we need to note an identity:

$sin2x = 2sinxcosx$

Using the identity from above, rewrite the equation.

$2sinxcosx - cosx=0$

Now factor out a $cosx$ .

$cosx(2sinx - 1) = 0$

Now take each factor and set it equal to zero.

$cosx = 0$

$x = pi/2, (3pi)/2$

$2sinx-1 = 0$

$2sinx = 1$

$sinx = 1/2$

$x = pi/6, (5pi)/6$

How do you solve sin2x-cosx=0sin2x-cosx=0?