Find all solutions to the equation in the interval [0,2). The equation being sin2x = cosx

sin2x-cosx

#1

Find all solutions to the equation in the interval [0,2). The equation being sin2x = cosx

Answer:

Assuming the interval is [0, 2pi)…
First, replace sin2x with the double angle formula: sin2x = 2sinx cosx
2sinx cosx = cosx
Move cosx to left side of equation:
2sinxcosx - cosx = 0
Factor out common factor of cos x
cos x (2sinx - 1) = 0
Use the Zero Product Property and solve for x
cos x = 0 OR 2 sinx - 1 = 0
If cos x = 0, then x = pi/2 OR x = 3pi/2
If 2 sinx - 1 = 0, then sinx = ½, so x = pi/6 OR x = 5 pi/6 (first and second quadrants)