What is the general solution for tan^2(x)-1=4cos(x) ?

The general solution for tan^2(x)-1=4cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin

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tan^2(x)-1=4cos(x)

tan^{2} (x) - 1 = 4 cos (x)

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

Solution steps

tan^{2} (x) - 1 = 4 cos (x)

Square both sides

(tan^{2} (x) - 1)^{2} = (4 cos (x))^{2}

Subtract

(4 cos (x))^{2}

from both sides

(tan^{2} (x) - 1)^{2} - 16 cos^{2} (x) = 0

Rewrite using trig identities

(- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - 16 cos^{2} (x) = 0

Factor

(- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - 16 cos^{2} (x) : (- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 cos (x)) (- 1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 cos (x))

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

Solving each part separately

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

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

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

\frac{10}{sin ( 3 0 ^{\circ} )} = \frac{3}{sin ( b )}

tan (\frac{θ}{2} - \frac{π}{6}) = - 1, 0 \leq θ < 2 π

arctan (x + 2) - arctan (x + 1) = \frac{π}{4}

3 cos (2 x) = 1

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

What is the general solution for tan^2(x)-1=4cos(x) ?
The general solution for tan^2(x)-1=4cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin