What is the general solution for 4sin(θ)=sqrt(3)sec(θ),0<= θ<180 ?

The general solution for 4sin(θ)=sqrt(3)sec(θ),0<= θ<180 is θ=30,θ=60

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4sin(θ)=sqrt(3)sec(θ),0<= θ<180

4 sin (θ) = 3 sec (θ), 0 \leq θ < 18 0^{\circ}

θ = 3 0^{\circ}, θ = 6 0^{\circ}

Solution steps

4 sin (θ) = 3 sec (θ), 0 \leq θ < 18 0^{\circ}

Subtract

3 sec (θ)

from both sides

4 sin (θ) - 3 sec (θ) = 0

Express with sin, cos

\frac{- 3 + 4 cos ( θ ) sin ( θ )}{cos ( θ )} = 0

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

- 3 + 4 cos (θ) sin (θ) = 0

Rewrite using trig identities

- 3 + 4 \cdot \frac{sin ( 2 θ )}{2} = 0

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

- 3 + 2 sin (2 θ) = 0

Move

3

to the right side

2 sin (2 θ) = 3

Divide both sides by

2

sin (2 θ) = \frac{3}{2}

General solutions for

sin (2 θ) = \frac{3}{2}

2 θ = 6 0^{\circ} + 36 0^{\circ} n, 2 θ = 12 0^{\circ} + 36 0^{\circ} n

Solve

2 θ = 6 0^{\circ} + 36 0^{\circ} n : θ = 3 0^{\circ} + 18 0^{\circ} n

Solve

2 θ = 12 0^{\circ} + 36 0^{\circ} n : θ = 6 0^{\circ} + 18 0^{\circ} n

θ = 3 0^{\circ} + 18 0^{\circ} n, θ = 6 0^{\circ} + 18 0^{\circ} n

Solutions for the range

0 \leq θ < 18 0^{\circ}

θ = 3 0^{\circ}, θ = 6 0^{\circ}

\frac{sin ( 8 2 ^{\circ} )}{sin ( x )} = 5

\frac{2 cos ^{2} ( x )}{2 ( 1 - sin ( x ) ) - cos ^{2} ( x )} = 0

2 sin (x) = - 3

tan (θ) = \frac{- 12}{5}, cot (θ)

3 cos (1 x) = - sin (1 x)

What is the general solution for 4sin(θ)=sqrt(3)sec(θ),0<= θ<180 ?
The general solution for 4sin(θ)=sqrt(3)sec(θ),0<= θ<180 is θ=30,θ=60