What is the general solution for 4sin(θ)= 1/(cos(θ)) ?

The general solution for 4sin(θ)= 1/(cos(θ)) is θ= pi/(12)+pin,θ=(5pi)/(12)+pin

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4sin(θ)= 1/(cos(θ))

4 sin (θ) = \frac{1}{cos ( θ )}

θ = \frac{π}{12} + πn, θ = \frac{5 π}{12} + πn

Solution steps

4 sin (θ) = \frac{1}{cos ( θ )}

Subtract

\frac{1}{cos ( θ )}

from both sides

4 sin (θ) - \frac{1}{cos ( θ )} = 0

Simplify

4 sin (θ) - \frac{1}{cos ( θ )} : \frac{4 sin ( θ ) cos ( θ ) - 1}{cos ( θ )}

\frac{4 sin ( θ ) cos ( θ ) - 1}{cos ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 sin (θ) cos (θ) - 1 = 0

Rewrite using trig identities

- 1 + 4 \cdot \frac{sin ( 2 θ )}{2} = 0

4 \cdot \frac{sin ( 2 θ )}{2} = 2 sin (2 θ)

- 1 + 2 sin (2 θ) = 0

Move

1

to the right side

2 sin (2 θ) = 1

Divide both sides by

2

sin (2 θ) = \frac{1}{2}

General solutions for

sin (2 θ) = \frac{1}{2}

2 θ = \frac{π}{6} + 2 πn, 2 θ = \frac{5 π}{6} + 2 πn

Solve

2 θ = \frac{π}{6} + 2 πn : θ = \frac{π}{12} + πn

Solve

2 θ = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{12} + πn

θ = \frac{π}{12} + πn, θ = \frac{5 π}{12} + πn

3 sin (2 x) = - sin (x)

10 sin (A) + 16 = 3 sin (A) + 9

y, x = cos (y)

tan^{2} (x) - 2 tan (x) - 5 = 0

tan (x) = \frac{10}{11}

What is the general solution for 4sin(θ)= 1/(cos(θ)) ?
The general solution for 4sin(θ)= 1/(cos(θ)) is θ= pi/(12)+pin,θ=(5pi)/(12)+pin