What is the general solution for 8sin(θ)cos(θ)=3 ?

The general solution for 8sin(θ)cos(θ)=3 is θ=(0.84806…)/2+pin,θ= pi/2-(0.84806…)/2+pin

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8sin(θ)cos(θ)=3

8 sin (θ) cos (θ) = 3

θ = \frac{0.84806 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.84806 \dots}{2} + πn

Solution steps

8 sin (θ) cos (θ) = 3

Rewrite using trig identities

8 \cdot \frac{sin ( 2 θ )}{2} = 3

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

\frac{sin ( 2 θ ) \cdot 8}{2} = 3

Divide the numbers:

\frac{8}{2} = 4

4 sin (2 θ) = 3

Divide both sides by

4

sin (2 θ) = \frac{3}{4}

Apply trig inverse properties

2 θ = arcsin (\frac{3}{4}) + 2 πn, 2 θ = π - arcsin (\frac{3}{4}) + 2 πn

Solve

2 θ = arcsin (\frac{3}{4}) + 2 πn : θ = \frac{arcsin ( \frac{3}{4} )}{2} + πn

Solve

2 θ = π - arcsin (\frac{3}{4}) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( \frac{3}{4} )}{2} + πn

θ = \frac{arcsin ( \frac{3}{4} )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( \frac{3}{4} )}{2} + πn

Show solutions in decimal form

θ = \frac{0.84806 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.84806 \dots}{2} + πn

tan (x) = \frac{7}{9}

cos (8 x) = - 1

- 3 cos^{2} (x) + 2 cos (x) + 1 = 0

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

cos (x + \frac{3 π}{2}) + 2 sin (x + π) = 0

What is the general solution for 8sin(θ)cos(θ)=3 ?
The general solution for 8sin(θ)cos(θ)=3 is θ=(0.84806…)/2+pin,θ= pi/2-(0.84806…)/2+pin