What is the general solution for sin(x)+cos(x)cot(x)=2 ?

The general solution for sin(x)+cos(x)cot(x)=2 is x= pi/6+2pin,x=(5pi)/6+2pin

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Popular >

sin(x)+cos(x)cot(x)=2

Solution

sin (x) + cos (x) cot (x) = 2

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (x) + cos (x) cot (x) = 2

Subtract

2

from both sides

sin (x) + cos (x) cot (x) - 2 = 0

Express with sin, cos

- 2 + sin (x) + cos (x) cot (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= - 2 sin (x) + 1

- 2 sin (x) + 1 = 0

Move

1

to the right side

- 2 sin (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

- 2 sin (x) = - 1

- 2 sin (x) = - 1

Divide both sides by

- 2

- 2 sin (x) = - 1

Divide both sides by

- 2

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

Simplify

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

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

General solutions for

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

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Popular Examples

2sin^2(x/2)+cos(2x)=0

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

tan(θ)+cot(θ)=2

tan (θ) + cot (θ) = 2

cos(x)= 6/20

cos (x) = \frac{6}{20}

2sin(x-60)=cos(x-30)

2 sin (x - 6 0^{\circ}) = cos (x - 3 0^{\circ})

2sin(θ)=2cos(θ)

2 sin (θ) = 2 cos (θ)

Frequently Asked Questions (FAQ)

What is the general solution for sin(x)+cos(x)cot(x)=2 ?
The general solution for sin(x)+cos(x)cot(x)=2 is x= pi/6+2pin,x=(5pi)/6+2pin