What is the general solution for 4sin^2(θ)-8cos(θ)+1=0 ?

The general solution for 4sin^2(θ)-8cos(θ)+1=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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4sin^2(θ)-8cos(θ)+1=0

4 sin^{2} (θ) - 8 cos (θ) + 1 = 0

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solution steps

4 sin^{2} (θ) - 8 cos (θ) + 1 = 0

Rewrite using trig identities

5 - 4 cos^{2} (θ) - 8 cos (θ) = 0

Solve by substitution

cos (θ) = - \frac{5}{2}, cos (θ) = \frac{1}{2}

cos (θ) = - \frac{5}{2} :

No Solution

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Combine all the solutions

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

sin (x) cos (\frac{π}{7}) - sin (\frac{π}{7}) cos (x) = \frac{2}{2}

20 sin (t) cos (t) = - 16 cos (t)

cos (x + 2 5^{\circ}) sec (6 5^{\circ} - x) = 1

2 cos^{2} (x) = 3 (sin (x) + 1)

cos^{2} (θ) = \frac{2 + 3}{4}

What is the general solution for 4sin^2(θ)-8cos(θ)+1=0 ?
The general solution for 4sin^2(θ)-8cos(θ)+1=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin