What is the general solution for 2sin(θ)=3cos(θ)-1,0<,θ<360 ?

The general solution for 2sin(θ)=3cos(θ)-1,0<,θ<360 is No Solution for θ\in\mathbb{R}

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2sin(θ)=3cos(θ)-1,0<,θ<360

2 sin (θ) = 3 cos (θ) - 1, 0 <, θ < 360

No Solution for θ \in R

Solution steps

2 sin (θ) = 3 cos (θ) - 1, 0, θ < 360

Square both sides

(2 sin (θ))^{2} = (3 cos (θ) - 1)^{2}

Subtract

(3 cos (θ) - 1)^{2}

from both sides

4 sin^{2} (θ) - 9 cos^{2} (θ) + 6 cos (θ) - 1 = 0

Rewrite using trig identities

3 - 13 cos^{2} (θ) + 6 cos (θ) = 0

Solve by substitution

cos (θ) = - \frac{- 3 + 4 3}{13}, cos (θ) = \frac{3 + 4 3}{13}

cos (θ) = - \frac{- 3 + 4 3}{13}, 0 :

No Solution

cos (θ) = \frac{3 + 4 3}{13}, 0 :

No Solution

Combine all the solutions

No Solution

Verify solutions by plugging them into the original equation

No Solution for θ \in R

cot (2 θ) + csc (2 θ) - 7 = 0

sec (x) sin (x) + cos (x) = sec (x)

0 = 2 + 2 sin (θ)

x, 2 cos (x) + 3 = 0

4 cos (x) - 5 = - 8

What is the general solution for 2sin(θ)=3cos(θ)-1,0<,θ<360 ?
The general solution for 2sin(θ)=3cos(θ)-1,0<,θ<360 is No Solution for θ\in\mathbb{R}