What is the general solution for 3=20.39sin(θ)cos(θ) ?

The general solution for 3=20.39sin(θ)cos(θ) is θ=(0.29868…)/2+pin,θ= pi/2-(0.29868…)/2+pin

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3=20.39sin(θ)cos(θ)

3 = 20.39 sin (θ) cos (θ)

θ = \frac{0.29868 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.29868 \dots}{2} + πn

Solution steps

3 = 20.39 sin (θ) cos (θ)

Switch sides

20.39 sin (θ) cos (θ) = 3

Rewrite using trig identities

20.39 \cdot \frac{sin ( 2 θ )}{2} = 3

Refine

20.39 \cdot \frac{sin ( 2 θ )}{2} : \frac{20.39 sin ( 2 θ )}{2}

\frac{20.39 sin ( 2 θ )}{2} = 3

Multiply both sides by

2

20.39 sin (2 θ) = 6

Divide both sides by

20.39

sin (2 θ) = 0.29426 \dots

Apply trig inverse properties

2 θ = arcsin (0.29426 \dots) + 2 πn, 2 θ = π - arcsin (0.29426 \dots) + 2 πn

Solve

2 θ = arcsin (0.29426 \dots) + 2 πn : θ = \frac{arcsin ( 0.29426 \dots )}{2} + πn

Solve

2 θ = π - arcsin (0.29426 \dots) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( 0.29426 \dots )}{2} + πn

θ = \frac{arcsin ( 0.29426 \dots )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( 0.29426 \dots )}{2} + πn

Show solutions in decimal form

θ = \frac{0.29868 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.29868 \dots}{2} + πn

sin^{4} (x) = \frac{1}{16}

sec (y) = \frac{4}{5}

(tan^{2} (x) - 1) (csc (x) + 1) = 0

12 sin^{2} (θ) - 3 = 0, 0 \leq θ \leq 2 π

0 = - 7.5 sin (\frac{π}{10} (t - 5)) + 8

What is the general solution for 3=20.39sin(θ)cos(θ) ?
The general solution for 3=20.39sin(θ)cos(θ) is θ=(0.29868…)/2+pin,θ= pi/2-(0.29868…)/2+pin