How do you solve $cos2x = sinx$ on the interval $0 <= x<= 2pi$ ?

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Answer 1 · 2015-04-16T14:56:16

If $cos(2x) = sin(x)$
then
$1-2sin^2(x) = sin(x)$

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

Substituting $k=sin(x)$
$2k^2+k-1 = 0$

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

$sin(x) = 1/2$ or $sin(x) =-1$

If $sin(x) = 1/2$ (for $0<=x<=2pi$ )
$x=pi/6 =30^o$ or $x= (5pi)/6 = 150^o$

If $sin(x) = -1$ (for $0<=x<=2pi$ )
$x=(3pi)/2 = 270^o$

So $x epsilon {pi/6, (5pi)/6, (3pi)/2}$
(or their equivalent in degrees)

How do you solve cos2x = sinxcos2x = sinx on the interval 0 <= x<= 2pi0 <= x<= 2pi?