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

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

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4sin^2(θ)-8cos(θ)+1=0,0<= θ<2pi

4 sin^{2} (θ) - 8 cos (θ) + 1 = 0, 0 \leq θ < 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

Solution steps

4 sin^{2} (θ) - 8 cos (θ) + 1 = 0, 0 \leq θ < 2 π

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}, 0 \leq θ < 2 π :

No Solution

cos (θ) = \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{π}{3}, θ = \frac{5 π}{3}

Combine all the solutions

θ = \frac{π}{3}, θ = \frac{5 π}{3}

4 sin (A) + 3 = 2 sin (A) + 4

0 = - 18 cos (3 x)

cos^{2} (2 x) - 3 cos (2 x) = 0

csc^{2} (x) = sin (x)

cos (θ) = \frac{4}{7}, tan (θ) < 0

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