What is the general solution for cos(2x)sin(x)+sin(x)=5cos(2x)+5 ?

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

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Popular >

cos(2x)sin(x)+sin(x)=5cos(2x)+5

Solution

cos (2 x) sin (x) + sin (x) = 5 cos (2 x) + 5

Solution

x = \frac{π}{2} + πn

Degrees

x = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

cos (2 x) sin (x) + sin (x) = 5 cos (2 x) + 5

Subtract

5 cos (2 x) + 5

from both sides

cos (2 x) sin (x) + sin (x) - 5 cos (2 x) - 5 = 0

Factor

cos (2 x) sin (x) + sin (x) - 5 cos (2 x) - 5 : (cos (2 x) + 1) (sin (x) - 5)

cos (2 x) sin (x) + sin (x) - 5 cos (2 x) - 5

= (sin (x) cos (2 x) + sin (x)) + (- 5 cos (2 x) - 5)

Factor out

- 5

from

- 5 cos (2 x) - 5 : - 5 (cos (2 x) + 1)

- 5 cos (2 x) - 5

Factor out common term

- 5

= - 5 (cos (2 x) + 1)

Factor out

sin (x)

from

sin (x) cos (2 x) + sin (x) : sin (x) (cos (2 x) + 1)

sin (x) cos (2 x) + sin (x)

Factor out common term

sin (x)

= sin (x) (cos (2 x) + 1)

= - 5 (cos (2 x) + 1) + sin (x) (cos (2 x) + 1)

Factor out common term

cos (2 x) + 1

= (cos (2 x) + 1) (sin (x) - 5)

(cos (2 x) + 1) (sin (x) - 5) = 0

Solving each part separately

cos (2 x) + 1 = 0 or sin (x) - 5 = 0

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

cos (2 x) + 1 = 0

Move

1

to the right side

cos (2 x) + 1 = 0

Subtract

1

from both sides

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

Simplify

cos (2 x) = - 1

cos (2 x) = - 1

General solutions for

cos (2 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}

2 x = π + 2 πn

2 x = π + 2 πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

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

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{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} + πn

Popular Examples

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3 cot (θ) = 3

sec(x)=1,-2pi<= x<= 2pi

sec (x) = 1, - 2 π \leq x \leq 2 π

tan(t)= 4/3 ,sin(t)

tan (t) = \frac{4}{3}, sin (t)

cos(x)= 2/(sqrt(6))

cos (x) = \frac{2}{6}

16sin^2(x)=9

16 sin^{2} (x) = 9

Frequently Asked Questions (FAQ)

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