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

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1 Answer

(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 √