(sin2x)(sinx) + cosx = 0 find all solutions of the equation in the interval [0,2pi]

Question

(sin2x)(sinx) + cosx = 0

Answer 1

(sin2x)(sinx) + cosx = 0

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

cos(x) (2 sin^2(x) + 1) = 0

Using zero product rule:

cos(x) = 0  or  2 sin^2(x) + 1 = 0

cos(x) = 0 when the unit circle intersects the vertical axis at angles of x = pi/2 and x = 3pi/2.

2 sin^2(x) + 1 = 0

sin^2(x) = -1/2

sin(x) = +-sqrt(-1/2) which has no solution b/c the values of sin(x) are all real.

 

So the answers are x = pi/2 and x = 3 pi/2

 

Check: (sin2x)(sinx) + cosx =? 0

Case 1: x = pi/2 => (sin pi)(sin pi/2) + cos pi/2 =

                               0 * 1 + 0 = 0 √

Case 2: x = 3 pi/2 => (sin(3pi))(sin(3 pi/2)) + cos(3 pi/2) =

                                  0 * (-1) + 0 = 0 √