What is the general solution for tan^2(θ)=-3/2 sec(θ),0<= θ<2pi ?

The general solution for tan^2(θ)=-3/2 sec(θ),0<= θ<2pi is θ=(2pi)/3 ,θ=(4pi)/3

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tan^2(θ)=-3/2 sec(θ),0<= θ<2pi

tan^{2} (θ) = - \frac{3}{2} sec (θ), 0 \leq θ < 2 π

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Solution steps

tan^{2} (θ) = - \frac{3}{2} sec (θ), 0 \leq θ < 2 π

Subtract

- \frac{3}{2} sec (θ)

from both sides

tan^{2} (θ) + \frac{3}{2} sec (θ) = 0

Simplify

tan^{2} (θ) + \frac{3}{2} sec (θ) : \frac{2 tan ^{2} ( θ ) + 3 sec ( θ )}{2}

\frac{2 tan ^{2} ( θ ) + 3 sec ( θ )}{2} = 0

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

2 tan^{2} (θ) + 3 sec (θ) = 0

Rewrite using trig identities

(- 1 + sec^{2} (θ)) \cdot 2 + 3 sec (θ) = 0

Solve by substitution

sec (θ) = \frac{1}{2}, sec (θ) = - 2

sec (θ) = \frac{1}{2}, 0 \leq θ < 2 π :

No Solution

sec (θ) = - 2, 0 \leq θ < 2 π : θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Combine all the solutions

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

5 sin (x) + 3 cos (x) = 0

3 cos (x) csc (x) = 2 cos (x)

sin (2 x) = - \frac{2}{2}

sin (x) = 0, 0 \leq x < 2 π

sec^{2} (x) = (3 + 1) tan (x) - 3 + 1

What is the general solution for tan^2(θ)=-3/2 sec(θ),0<= θ<2pi ?
The general solution for tan^2(θ)=-3/2 sec(θ),0<= θ<2pi is θ=(2pi)/3 ,θ=(4pi)/3