What is the general solution for 2+7cos(θ)=4sin^2(θ) ?

The general solution for 2+7cos(θ)=4sin^2(θ) is θ=1.31811…+2pin,θ=2pi-1.31811…+2pin

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2+7cos(θ)=4sin^2(θ)

2 + 7 cos (θ) = 4 sin^{2} (θ)

θ = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn

Solution steps

2 + 7 cos (θ) = 4 sin^{2} (θ)

Subtract

4 sin^{2} (θ)

from both sides

2 + 7 cos (θ) - 4 sin^{2} (θ) = 0

Rewrite using trig identities

- 2 + 4 cos^{2} (θ) + 7 cos (θ) = 0

Solve by substitution

cos (θ) = \frac{1}{4}, cos (θ) = - 2

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

cos (θ) = - 2 :

No Solution

Combine all the solutions

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

Show solutions in decimal form

θ = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn

x, sinh (x) = 0

sin (a) = 0.5

2 sin^{4} (x) = cos^{2} (x)

2 cos (3 x) = cot (3 x)

cos (x) (cos (x) - 1) = 0, n = 0

What is the general solution for 2+7cos(θ)=4sin^2(θ) ?
The general solution for 2+7cos(θ)=4sin^2(θ) is θ=1.31811…+2pin,θ=2pi-1.31811…+2pin