Question #77b55

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Answer 1 · 2016-11-19T17:26:12

$sinxcosx = 0 -> (sinxcosx)^2 = 0^2$

$sin^2xcos^2x = 0$

Use the identity $sin^2theta + cos^2theta =1-> cos^2theta = 1 -sin^2theta$ .

$sin^2x(1 - sin^2x) = 0$

$sin^2x = 0 and sin^2x = 1$

$sinx = 0 and sinx = +-1$

$x = 0 + 2pin, pi + 2pin, pi/2 + 2pin, (3pi)/2 + 2pin$

Hopefully this helps!

Answer 2 · 2017-09-18T03:28:10

$x = pin, pi/2 + pin$

Alternatively, we have:

$2sinxcosx = sin(2x)$ .

$1/2sin(2x) = 0$

$sin(2x) = 0$

$2x = 0, pi$

$x = 0 + pin, pi/2 + pin$

Hopefully this helps!