What is the general solution for 2sin(x)+cot(x)-csc(x)=0 ?

The general solution for 2sin(x)+cot(x)-csc(x)=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

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2sin(x)+cot(x)-csc(x)=0

2 sin (x) + cot (x) - csc (x) = 0

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

Solution steps

2 sin (x) + cot (x) - csc (x) = 0

Express with sin, cos

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

Simplify

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

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

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

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

Subtract

cos (x)

from both sides

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

Square both sides

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

Subtract

(- cos (x))^{2}

from both sides

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

Factor

(2 sin^{2} (x) - 1)^{2} - cos^{2} (x) : (2 sin^{2} (x) - 1 + cos (x)) (2 sin^{2} (x) - 1 - cos (x))

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

Solving each part separately

2 sin^{2} (x) - 1 + cos (x) = 0 or 2 sin^{2} (x) - 1 - cos (x) = 0

2 sin^{2} (x) - 1 + cos (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2 πn

2 sin^{2} (x) - 1 - cos (x) = 0 : x = π + 2 πn, 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 = 2 πn, x = π + 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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cosh (x) = \frac{5}{3}

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

sin (θ) = 0.75

x, sin (x) + cos (x) = 1

5 cos (x) - 5 = 0

What is the general solution for 2sin(x)+cot(x)-csc(x)=0 ?
The general solution for 2sin(x)+cot(x)-csc(x)=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin