What is the general solution for sin(x)tan(x)+1=sin(x)+tan(x) ?

The general solution for sin(x)tan(x)+1=sin(x)+tan(x) is x= pi/4+pin

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sin(x)tan(x)+1=sin(x)+tan(x)

Solution

sin (x) tan (x) + 1 = sin (x) + tan (x)

Solution

x = \frac{π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

sin (x) tan (x) + 1 = sin (x) + tan (x)

Subtract

sin (x) + tan (x)

from both sides

sin (x) tan (x) + 1 - sin (x) - tan (x) = 0

Express with sin, cos

1 - sin (x) - tan (x) + sin (x) tan (x)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

Simplify

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

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

sin (x) \frac{sin ( x )}{cos ( x )} = \frac{sin ^{2} ( x )}{cos ( x )}

sin (x) \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

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

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

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (x) sin (x) = sin^{1 + 1} (x)

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

= \frac{sin ^{2} ( x )}{cos ( x )}

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

Combine the fractions

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

Apply rule

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

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

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

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

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

Factor

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

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

Factor out common term

cos (x)

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

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

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

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

Factor out common term

sin (x)

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

Rewrite as

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

Factor out common term

(1 - sin (x))

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

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

Solving each part separately

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

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

1 - sin (x) = 0

Move

1

to the right side

1 - sin (x) = 0

Subtract

1

from both sides

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

Simplify

- sin (x) = - 1

- sin (x) = - 1

Divide both sides by

- 1

- sin (x) = - 1

Divide both sides by

- 1

\frac{- sin ( x )}{- 1} = \frac{- 1}{- 1}

Simplify

sin (x) = 1

sin (x) = 1

General solutions for

sin (x) = 1

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 = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

cos (x) - sin (x) = 0 : x = \frac{π}{4} + πn

cos (x) - sin (x) = 0

Rewrite using trig identities

cos (x) - sin (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 - tan (x) = 0

1 - tan (x) = 0

Move

1

to the right side

1 - tan (x) = 0

Subtract

1

from both sides

1 - tan (x) - 1 = 0 - 1

Simplify

- tan (x) = - 1

- tan (x) = - 1

Divide both sides by

- 1

- tan (x) = - 1

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{- 1}{- 1}

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{π}{4} + πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn

x = \frac{π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x)tan(x)+1=sin(x)+tan(x) ?
The general solution for sin(x)tan(x)+1=sin(x)+tan(x) is x= pi/4+pin