What is the general solution for 2cos^2(θ)+sin(θ)=2 ?

The general solution for 2cos^2(θ)+sin(θ)=2 is θ=2pin,θ=pi+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin

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2cos^2(θ)+sin(θ)=2

2 cos^{2} (θ) + sin (θ) = 2

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

2 cos^{2} (θ) + sin (θ) = 2

Subtract

2

from both sides

2 cos^{2} (θ) + sin (θ) - 2 = 0

Rewrite using trig identities

sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 0, sin (θ) = \frac{1}{2}

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

tanh^{2} (x) + 5 sech (x) - 5 = 0

a, P = cot^{2} (a)

10 sin^{2} (2 u) + 6 cos^{2} (2 u) = 8

x, sin (x θ) = \frac{1}{2}

x, tan (x) = \frac{3.057}{6}

What is the general solution for 2cos^2(θ)+sin(θ)=2 ?
The general solution for 2cos^2(θ)+sin(θ)=2 is θ=2pin,θ=pi+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin