What is the general solution for sin^2(2θ)-5sin(2θ)-1=0 ?

The general solution for sin^2(2θ)-5sin(2θ)-1=0 is θ=(-0.19379…)/2+pin,θ= pi/2+(0.19379…)/2+pin

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sin^2(2θ)-5sin(2θ)-1=0

sin^{2} (2 θ) - 5 sin (2 θ) - 1 = 0

θ = \frac{- 0.19379 \dots}{2} + πn, θ = \frac{π}{2} + \frac{0.19379 \dots}{2} + πn

Solution steps

sin^{2} (2 θ) - 5 sin (2 θ) - 1 = 0

Solve by substitution

sin (2 θ) = \frac{5 + 29}{2}, sin (2 θ) = \frac{5 - 29}{2}

sin (2 θ) = \frac{5 + 29}{2} :

No Solution

sin (2 θ) = \frac{5 - 29}{2} : θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

Combine all the solutions

θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

Show solutions in decimal form

θ = \frac{- 0.19379 \dots}{2} + πn, θ = \frac{π}{2} + \frac{0.19379 \dots}{2} + πn

- \frac{π}{30} sin (\frac{π}{90} (x + 20)) = 0

sin (2 x) = - \frac{1}{2}

cos (x) = - \frac{22}{7.28 \cdot 20}

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

x, 1 + sin (2 x - 3 0^{\circ}) = 0

What is the general solution for sin^2(2θ)-5sin(2θ)-1=0 ?
The general solution for sin^2(2θ)-5sin(2θ)-1=0 is θ=(-0.19379…)/2+pin,θ= pi/2+(0.19379…)/2+pin