What is the general solution for 1/(cos(x))+sec^2(x)-1-cos(x)=0 ?

The general solution for 1/(cos(x))+sec^2(x)-1-cos(x)=0 is x=pi+2pin,x=2pin

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

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

x = π + 2 πn, x = 2 πn

Solution steps

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

Simplify

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

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

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

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

Rewrite using trig identities

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

Solve by substitution

sec (x) = - 1, sec (x) = 1

sec (x) = - 1 : x = π + 2 πn

sec (x) = 1 : x = 2 πn

Combine all the solutions

x = π + 2 πn, x = 2 πn

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

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

tan (θ) = \frac{3}{3}, 18 0^{\circ} \leq θ \leq 36 0^{\circ}

7.5 = 5 sin (\frac{π}{2} (x - 2)) + 5

u, y = ccos h \cdot sin (u)

What is the general solution for 1/(cos(x))+sec^2(x)-1-cos(x)=0 ?
The general solution for 1/(cos(x))+sec^2(x)-1-cos(x)=0 is x=pi+2pin,x=2pin