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

The general solution for sec(x)sin(x)+cos(x)=sec(x) is x=2pin,x=pi+2pin

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sec(x)sin(x)+cos(x)=sec(x)

sec (x) sin (x) + cos (x) = sec (x)

x = 2 πn, x = π + 2 πn

Solution steps

sec (x) sin (x) + cos (x) = sec (x)

Subtract

sec (x)

from both sides

sec (x) sin (x) + cos (x) - sec (x) = 0

Express with sin, cos

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

Simplify

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

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

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

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

Subtract

cos^{2} (x)

from both sides

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

Square both sides

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

Subtract

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

from both sides

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

Factor

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

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

Solving each part separately

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

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

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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

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

Since the equation is undefined for:

\frac{π}{2} + 2 πn

x = 2 πn, x = π + 2 πn

0 = 2 + 2 sin (θ)

x, 2 cos (x) + 3 = 0

4 cos (x) - 5 = - 8

sin (x) + 4 csc (x) + 5 = 0, 0 \leq x \leq 2 π

r = a sin (3 x)

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