Best Answer: sin2x sinx - cosx=0 --> [2sin(x)cos(x)]sin(x) - cos(x) = 0 --> 2sin^2(x)cos(x) - cos(x) = 0 --> 2[1 - cos^2(x)]cos(x) - cos(x) = 0 --> 2cos(x) - 2cos^3(x) - cos(x) = 0 --> 2cos^3(x) + cos(x) = 0 Let y = cos(x), then 2y^3 + y = 0 --> y(2y^2 + 1) = 0 --> y = 0 --> cos(x) = 0 i.e. x = (2k+1)pi/2 for k = ..., -1, 0, 1, ....

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up