Answered

Get the information you need with the help of IDNLearn.com's expert community. Join our interactive community and get comprehensive, reliable answers to all your questions.

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

Sagot :

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)

Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Your questions find answers at IDNLearn.com. Thanks for visiting, and come back for more accurate and reliable solutions.