Question

Find the exact solutions of the equation in the interval [0, 2pi). (Enter your answers as a comma-separated list.)

sin 2x sin x = COS X

Answer 1

there are different ways of solving this:

one way is the follwing:

cos^2(x) -sin^2(x) =cos (2x)

then, equation becomes:

cos (2x) -sin (x)=0

but sin(x) is also equivalent to : cos ((pi/2)-x),

so, the equation now is:

cos (2x) - cos ((pi/2)-x)=0 -------> cos (2x) = cos ((pi/2)-x) ---------> 2x=(pi/2) -x

but since it is an angle, the answer would be repeating so:

2x= 2k(pi) +((pi/2)-x)

3x= pi( 2k+1/2)

x= (pi/3) (2k+1/2)

for k=0, x= pi/6

for k=1, x= 5pi/6

for k=2, x= 3pi/2

so answers are : (pi/6,5pi/6,3pi/2)