What is the general solution for tan(pi/2-x)+tan(x)-1=0 ?

The general solution for tan(pi/2-x)+tan(x)-1=0 is No Solution for x\in\mathbb{R}

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Popular >

tan(pi/2-x)+tan(x)-1=0

Solution

tan (\frac{π}{2} - x) + tan (x) - 1 = 0

Solution

No Solution for x \in R

Solution steps

tan (\frac{π}{2} - x) + tan (x) - 1 = 0

Rewrite using trig identities

tan (\frac{π}{2} - x) + tan (x) - 1 = 0

Rewrite using trig identities

tan (\frac{π}{2} - x)

Use the basic trigonometric identity:

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

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

Use the Angle Difference identity:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

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

Use the Angle Difference identity:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

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

Simplify

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

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

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

sin (\frac{π}{2}) cos (x) - cos (\frac{π}{2}) sin (x)

sin (\frac{π}{2}) cos (x) = cos (x)

sin (\frac{π}{2}) cos (x)

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 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}

= 1

= 1 \cdot cos (x)

Multiply:

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

= cos (x)

cos (\frac{π}{2}) sin (x) = 0

cos (\frac{π}{2}) sin (x)

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

= cos (x) - 0

cos (x) - 0 = cos (x)

= cos (x)

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

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

cos (\frac{π}{2}) cos (x) + sin (\frac{π}{2}) sin (x)

cos (\frac{π}{2}) cos (x) = 0

cos (\frac{π}{2}) cos (x)

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (x) = sin (x)

sin (\frac{π}{2}) sin (x)

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 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}

= 1

= 1 \cdot sin (x)

Multiply:

1 \cdot sin (x) = sin (x)

= sin (x)

= 0 + sin (x)

0 + sin (x) = sin (x)

= sin (x)

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

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

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

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

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

Express with sin, cos

cos (x) - sin (x) + sin (x) tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

cos (x) - sin (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 )}

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

Convert element to fraction:

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

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

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

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

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

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

Rewrite using trig identities

- cos (x) sin (x) + 1

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

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

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

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

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

Multiply both sides by

2

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

Multiply both sides by

2

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

Simplify

- sin (2 x) = - 2

- sin (2 x) = - 2

Divide both sides by

- 1

- sin (2 x) = - 2

Divide both sides by

- 1

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

Simplify

sin (2 x) = 2

sin (2 x) = 2

- 1 \leq sin (x) \leq 1

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for tan(pi/2-x)+tan(x)-1=0 ?
The general solution for tan(pi/2-x)+tan(x)-1=0 is No Solution for x\in\mathbb{R}