What is the general solution for 2tan(x)=3csc(x) ?

The general solution for 2tan(x)=3csc(x) is x= pi/3+2pin,x=(5pi)/3+2pin

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2tan(x)=3csc(x)

2 tan (x) = 3 csc (x)

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solution steps

2 tan (x) = 3 csc (x)

Subtract

3 csc (x)

from both sides

2 tan (x) - 3 csc (x) = 0

Express with sin, cos

2 \cdot \frac{sin ( x )}{cos ( x )} - 3 \cdot \frac{1}{sin ( x )} = 0

Simplify

2 \cdot \frac{sin ( x )}{cos ( x )} - 3 \cdot \frac{1}{sin ( x )} : \frac{2 sin ^{2} ( x ) - 3 cos ( x )}{cos ( x ) sin ( x )}

\frac{2 sin ^{2} ( x ) - 3 cos ( x )}{cos ( x ) sin ( x )} = 0

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

2 sin^{2} (x) - 3 cos (x) = 0

Add

3 cos (x)

to both sides

2 sin^{2} (x) = 3 cos (x)

Square both sides

(2 sin^{2} (x))^{2} = (3 cos (x))^{2}

Subtract

(3 cos (x))^{2}

from both sides

4 sin^{4} (x) - 9 cos^{2} (x) = 0

Factor

4 sin^{4} (x) - 9 cos^{2} (x) : (2 sin^{2} (x) + 3 cos (x)) (2 sin^{2} (x) - 3 cos (x))

(2 sin^{2} (x) + 3 cos (x)) (2 sin^{2} (x) - 3 cos (x)) = 0

Solving each part separately

2 sin^{2} (x) + 3 cos (x) = 0 or 2 sin^{2} (x) - 3 cos (x) = 0

2 sin^{2} (x) + 3 cos (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

2 sin^{2} (x) - 3 cos (x) = 0 : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Verify solutions by plugging them into the original equation

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

sin (t) = \frac{120}{500} sin (11 0^{\circ})

tan (θ) + 1 = 3 + 3 cot (θ)

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

3 = 4 sin (8 (- \frac{π}{2} - x)) + 3

5 sin (x) - 7 = 0

What is the general solution for 2tan(x)=3csc(x) ?
The general solution for 2tan(x)=3csc(x) is x= pi/3+2pin,x=(5pi)/3+2pin