What is the general solution for 5cot(x)=cos(x) ?

The general solution for 5cot(x)=cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin

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Popular >

5cot(x)=cos(x)

Solution

5 cot (x) = cos (x)

Solution

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

5 cot (x) = cos (x)

Subtract

cos (x)

from both sides

5 cot (x) - cos (x) = 0

Express with sin, cos

- cos (x) + 5 cot (x)

Use the basic trigonometric identity:

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

= - cos (x) + 5 \cdot \frac{cos ( x )}{sin ( x )}

Simplify

- cos (x) + 5 \cdot \frac{cos ( x )}{sin ( x )} : \frac{- cos ( x ) sin ( x ) + 5 cos ( x )}{sin ( x )}

- cos (x) + 5 \cdot \frac{cos ( x )}{sin ( x )}

Multiply

5 \cdot \frac{cos ( x )}{sin ( x )} : \frac{5 cos ( x )}{sin ( x )}

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

Multiply fractions:

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

= \frac{cos ( x ) \cdot 5}{sin ( x )}

= - cos (x) + \frac{5 cos ( x )}{sin ( x )}

Convert element to fraction:

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

= - \frac{cos ( x ) sin ( x )}{sin ( x )} + \frac{cos ( x ) \cdot 5}{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 ) sin ( x ) + cos ( x ) \cdot 5}{sin ( x )}

= \frac{- cos ( x ) sin ( x ) + 5 cos ( x )}{sin ( x )}

\frac{5 cos ( x ) - cos ( x ) sin ( x )}{sin ( x )} = 0

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

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

Factor

5 cos (x) - cos (x) sin (x) : - cos (x) (sin (x) - 5)

5 cos (x) - cos (x) sin (x)

Factor out common term

- cos (x)

= - cos (x) (- 5 + sin (x))

- cos (x) (sin (x) - 5) = 0

Solving each part separately

cos (x) = 0 or sin (x) - 5 = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 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}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

sin (x) - 5 = 0 :

No Solution

sin (x) - 5 = 0

Move

5

to the right side

sin (x) - 5 = 0

Add

5

to both sides

sin (x) - 5 + 5 = 0 + 5

Simplify

sin (x) = 5

sin (x) = 5

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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9 sin^{2} (θ) + 19 sin (θ) + 8 = 7 sin (θ) + 4

arccos(x)=(7pi)/8

arccos (x) = \frac{7 π}{8}

(sin(37))/(23)=(sin(x))/(21)

\frac{sin ( 3 7 ^{\circ} )}{23} = \frac{sin ( x )}{21}

csc(x)= 1/3

csc (x) = \frac{1}{3}

Frequently Asked Questions (FAQ)

What is the general solution for 5cot(x)=cos(x) ?
The general solution for 5cot(x)=cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin