What is the general solution for sin(2θ)= a/5 ?

The general solution for sin(2θ)= a/5 is θ=(arcsin(a/5))/2+pin,θ= pi/2+(arcsin(-a/5))/2+pin

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sin(2θ)= a/5

sin (2 θ) = \frac{a}{5}

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

Solution steps

sin (2 θ) = \frac{a}{5}

Apply trig inverse properties

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

Solve

2 θ = arcsin (\frac{a}{5}) + 2 πn : θ = \frac{arcsin ( \frac{a}{5} )}{2} + πn

Solve

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

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

cos (a) = - \frac{1}{4}

\frac{cos ^{2} ( x ) - 4}{cos ( x ) - 2} = 1

- 8 sec^{2} (x) - 26 tan (x) = 12

tan (θ) = 0.3

tan (θ) = 0.8

What is the general solution for sin(2θ)= a/5 ?
The general solution for sin(2θ)= a/5 is θ=(arcsin(a/5))/2+pin,θ= pi/2+(arcsin(-a/5))/2+pin