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

Solution

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

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

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 ) + cos ( x )}{cos ( x )} - sin (x) - 1

Convert element to fraction:

sin (x) = \frac{sin ( x ) cos ( x )}{cos ( x )}, 1 = \frac{1 cos ( x )}{cos ( x )}

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Add similar elements:

cos (x) - 1 \cdot cos (x) = 0

= \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) - sin (x) cos (x)

Factor out common term

- sin (x)

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

- 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

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

cos (x) - 1 = 0 : x = 2 πn

cos (x) - 1 = 0

Move

1

to the right side

cos (x) - 1 = 0

Add

1

to both sides

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

Simplify

cos (x) = 1

cos (x) = 1

General solutions for

cos (x) = 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

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

Frequently Asked Questions (FAQ)

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