Question

Find all values of x in the interval [0, 2π] that satisfy the equation.
3 sin(2x) = 3 cos(x)

Answer 1

sin 2x = 2 sinx cos x so:

3* 2 sinx cos x - 3 cos x = 0
3cos x ( 2 sin x - 1) = 0
cos x = 0 or 2sin x - 1 = 0

cos x = 0 or sin x = 1/2

for cos x = 0, one result is x = pi/2
for sin x = 1/2 one result is x = pi/6

can you find the others?

Answer 2

@cwrw238

great job but it would appreciated if you have the step
\[x= \cos^{-1}(0), and then say x = \pi/2, 3\pi/2,... \] as cos is 0 for pi/2 and 3pi/2
similarly
\[x=\sin^{-1} (1/2)=\pi/6,(5\pi/6),...\] as SIN is positive in second quadrant

Answer 3

HELLO you have
cos(pi/2-2x)=sin(2x)......

Answer 4

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