6cos^2(θ)-5=6sin^2(θ)+sin(θ)

Solution

6 cos^{2} (θ) - 5 = 6 sin^{2} (θ) + sin (θ)

θ = - 0.33983 \dots + 2 πn, θ = π + 0.33983 \dots + 2 πn, θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

Solution steps

6 cos^{2} (θ) - 5 = 6 sin^{2} (θ) + sin (θ)

Subtract

6 sin^{2} (θ) + sin (θ)

from both sides

6 cos^{2} (θ) - 5 - 6 sin^{2} (θ) - sin (θ) = 0

Rewrite using trig identities

1 - sin (θ) - 12 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = - \frac{1}{3}, sin (θ) = \frac{1}{4}

sin (θ) = - \frac{1}{3} : θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn

sin (θ) = \frac{1}{4} : θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

Combine all the solutions

θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn, θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

Show solutions in decimal form

θ = - 0.33983 \dots + 2 πn, θ = π + 0.33983 \dots + 2 πn, θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

sin (\frac{1}{3} x) = 0

2 sec^{2} (x) - 2 sec (x) = 4

sin (2 x) = - sin (4 x)

tan (x) + sin (x) = 0

cos (x) (2 sin (x) + 1) = 0, 0 \leq x \leq 2 π