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

Solution

sin (x) (2 cos^{2} (x) - 1) = 0

x = 2 πn, x = π + 2 πn, x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

Solution steps

sin (x) (2 cos^{2} (x) - 1) = 0

Solving each part separately

sin (x) = 0 or 2 cos^{2} (x) - 1 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

2 cos^{2} (x) - 1 = 0 : x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn, x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn, x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

Show solutions in decimal form

x = 2 πn, x = π + 2 πn, x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

tan (θ) = \frac{1}{8}

4 sin^{2} (x) + 2 sin (x) - 1 = 0

tan (θ) = \frac{1}{6}

cos (x) = - 12

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