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

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

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sec(x)+sin^2(x)+cos^2(x)=tan^2(x)

sec (x) + sin^{2} (x) + cos^{2} (x) = tan^{2} (x)

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

Solution steps

sec (x) + sin^{2} (x) + cos^{2} (x) = tan^{2} (x)

Subtract

tan^{2} (x)

from both sides

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

Express with sin, cos

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

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

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

Factor

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

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

Solving each part separately

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

1 + cos (x) = 0 : x = π + 2 πn

cos (x) (cos^{2} (x) + 1 - cos (x)) + sin^{2} (x) (- 1 + cos (x)) = 0 : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

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

4 sin^{2} (x) - 3 = 4, sin (2 x) - 3 = 0

tan^{2} (θ) + tan (θ) = 6, 0 \leq θ < 18 0^{\circ}

sin (2 x) = 5 sin (x)

2 cos^{2} (θ) - 4 cos (θ) - 1 = - 5 cos (θ)

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

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