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

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

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1/(2cos^2(x-1))=(1+tan^2(x))/(2sec^2(x))

\frac{1}{2 cos ^{2} ( x - 1 )} = \frac{1 + tan ^{2} ( x )}{2 sec ^{2} ( x )}

x = π + 2 πn + 1, x = 2 πn + 1

Solution steps

\frac{1}{2 cos ^{2} ( x - 1 )} = \frac{1 + tan ^{2} ( x )}{2 sec ^{2} ( x )}

Subtract

\frac{1 + tan ^{2} ( x )}{2 sec ^{2} ( x )}

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

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

Factor

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

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

Solving each part separately

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

sec^{2} (x) = 0 :

No Solution

cos (- 1 + x) + 1 = 0 : x = π + 2 πn + 1

cos (- 1 + x) - 1 = 0 : x = 2 πn + 1

Combine all the solutions

x = π + 2 πn + 1, x = 2 πn + 1

sin (2 x) - 2 cos (2 x) = 0

tan (x) = \frac{87.092}{86.676}

csc (x) = - \frac{3}{2}

1 + 3 sin (θ) = 0

\frac{tan ( 2 x )}{sin ( x )} = - 2

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