What is the general solution for 2sin(x)tan(x)+5=4sec(x) ?

The general solution for 2sin(x)tan(x)+5=4sec(x) is x= pi/3+2pin,x=(5pi)/3+2pin

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2sin(x)tan(x)+5=4sec(x)

2 sin (x) tan (x) + 5 = 4 sec (x)

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

Solution steps

2 sin (x) tan (x) + 5 = 4 sec (x)

Subtract

4 sec (x)

from both sides

2 sin (x) tan (x) + 5 - 4 sec (x) = 0

Express with sin, cos

2 sin (x) \frac{sin ( x )}{cos ( x )} + 5 - 4 \cdot \frac{1}{cos ( x )} = 0

Simplify

2 sin (x) \frac{sin ( x )}{cos ( x )} + 5 - 4 \cdot \frac{1}{cos ( x )} : \frac{2 sin ^{2} ( x ) - 4 + 5 cos ( x )}{cos ( x )}

\frac{2 sin ^{2} ( x ) - 4 + 5 cos ( x )}{cos ( x )} = 0

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

2 sin^{2} (x) - 4 + 5 cos (x) = 0

Subtract

5 cos (x)

from both sides

2 sin^{2} (x) - 4 = - 5 cos (x)

Square both sides

(2 sin^{2} (x) - 4)^{2} = (- 5 cos (x))^{2}

Subtract

(- 5 cos (x))^{2}

from both sides

(2 sin^{2} (x) - 4)^{2} - 25 cos^{2} (x) = 0

Factor

(2 sin^{2} (x) - 4)^{2} - 25 cos^{2} (x) : (2 sin^{2} (x) - 4 + 5 cos (x)) (2 sin^{2} (x) - 4 - 5 cos (x))

(2 sin^{2} (x) - 4 + 5 cos (x)) (2 sin^{2} (x) - 4 - 5 cos (x)) = 0

Solving each part separately

2 sin^{2} (x) - 4 + 5 cos (x) = 0 or 2 sin^{2} (x) - 4 - 5 cos (x) = 0

2 sin^{2} (x) - 4 + 5 cos (x) = 0 : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

2 sin^{2} (x) - 4 - 5 cos (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Combine all the solutions

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

cos^{2} (x) - cos (x) - 2 = 0, 0 \leq x \leq 2 π

cos^{4} (x) - 1 = 0

sec (x) - cos (x) = sin (x)

2 sin (x) = 3, 0 \leq x < 2 π

5 csc^{2} (x) + 2.5 cot (x) = 25

What is the general solution for 2sin(x)tan(x)+5=4sec(x) ?
The general solution for 2sin(x)tan(x)+5=4sec(x) is x= pi/3+2pin,x=(5pi)/3+2pin