How do you solve $3cosx - sin2x = 0$ from 0 to 2pi?

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Answer 1 · 2018-05-07T02:21:20

$3cosx-sin2x=0$

$=>3cosx-2sinxcosx=0$

$=>cosx(3-2sinx)=0$

So $cosx=0 or sinx=3/2>1" not possible"$

When $cosx=0=cos(pi/2)=cos(2pi-pi/2)$

Hence $x=pi/2or(3pi)/2$

Answer 2 · 2018-05-07T02:24:55

Values inconsistent.

$3 cos x - sin 2x = 0$

$3 cos x = sin 2x$

$3 cancel(cos x) = 2 sin x cancel(cos x)$

$3 = 2 sin x$

$sin x = 3/2$

As sine can have a value only between -1 & +1, values given in the sum are inconsistent.

How do you solve 3cosx - sin2x = 03cosx - sin2x = 0 from 0 to 2pi?