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

The general solution for tan(x)sin(x)+cos(x)=1 is x=2pin

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Popular >

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

Solution

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

Solution

x = 2 πn

Degrees

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

Solution steps

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

Subtract

1

from both sides

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

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

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

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

1 \cdot cos (x)

Multiply:

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

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

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= - cos (x) + 1

- cos (x) + 1 = 0

Move

1

to the right side

- cos (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

- cos (x) = - 1

- cos (x) = - 1

Divide both sides by

- 1

- cos (x) = - 1

Divide both sides by

- 1

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

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Frequently Asked Questions (FAQ)

What is the general solution for tan(x)sin(x)+cos(x)=1 ?
The general solution for tan(x)sin(x)+cos(x)=1 is x=2pin