What is the general solution for ([sin(x)+cos(x)])/(cos(x))=sin(x)+1 ?

The general solution for ([sin(x)+cos(x)])/(cos(x))=sin(x)+1 is x=2pin,x=pi+2pin

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([sin(x)+cos(x)])/(cos(x))=sin(x)+1

\frac{[ sin ( x ) + cos ( x ) ]}{cos ( x )} = sin (x) + 1

x = 2 πn, x = π + 2 πn

Solution steps

\frac{[ sin ( x ) + cos ( x ) ]}{cos ( x )} = sin (x) + 1

Subtract

sin (x) + 1

from both sides

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

Simplify

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

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

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

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

Factor

sin (x) - sin (x) cos (x) : - sin (x) (cos (x) - 1)

- sin (x) (cos (x) - 1) = 0

Solving each part separately

sin (x) = 0 or cos (x) - 1 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

cos (x) - 1 = 0 : x = 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

cos (A) = 0.4639

csc (A) = \frac{9.43}{5}

cos (x) cos (3 5^{\circ}) = 1 - sin (x) sin (3 5^{\circ})

4 sin^{2} (x) + 2 sin (x) = 2, 0 \leq x \leq 2 π

3 tan (\frac{1}{3} x) = 8

What is the general solution for ([sin(x)+cos(x)])/(cos(x))=sin(x)+1 ?
The general solution for ([sin(x)+cos(x)])/(cos(x))=sin(x)+1 is x=2pin,x=pi+2pin