What is the general solution for sin(2θ)=(-2sqrt(6))/7 ?

The general solution for sin(2θ)=(-2sqrt(6))/7 is θ=-(0.77519…)/2+pin,θ= pi/2+(0.77519…)/2+pin

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sin(2θ)=(-2sqrt(6))/7

sin (2 θ) = \frac{- 2 6}{7}

θ = - \frac{0.77519 \dots}{2} + πn, θ = \frac{π}{2} + \frac{0.77519 \dots}{2} + πn

Solution steps

sin (2 θ) = \frac{- 2 6}{7}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

sin (2 θ) = - \frac{2 6}{7}

Apply trig inverse properties

2 θ = arcsin (- \frac{2 6}{7}) + 2 πn, 2 θ = π + arcsin (\frac{2 6}{7}) + 2 πn

Solve

2 θ = arcsin (- \frac{2 6}{7}) + 2 πn : θ = - \frac{arcsin ( \frac{2 6}{7} )}{2} + πn

Solve

2 θ = π + arcsin (\frac{2 6}{7}) + 2 πn : θ = \frac{π}{2} + \frac{arcsin ( \frac{2 6}{7} )}{2} + πn

θ = - \frac{arcsin ( \frac{2 6}{7} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( \frac{2 6}{7} )}{2} + πn

Show solutions in decimal form

θ = - \frac{0.77519 \dots}{2} + πn, θ = \frac{π}{2} + \frac{0.77519 \dots}{2} + πn

x, sin (x) = - \frac{7}{25}

\frac{1}{tan ( α )} + tan (α) = \frac{1}{sin ( α )}

tan (2 θ) = \frac{- 81}{81}

sin (θ) cos (2 θ) - cos (2 θ) = 0

tan (θ) = - \frac{7}{6}

What is the general solution for sin(2θ)=(-2sqrt(6))/7 ?
The general solution for sin(2θ)=(-2sqrt(6))/7 is θ=-(0.77519…)/2+pin,θ= pi/2+(0.77519…)/2+pin