What is the general solution for sin(2x)+sin(x)+2cos(x)+1=0 ?

The general solution for sin(2x)+sin(x)+2cos(x)+1=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin,x=(3pi)/2+2pin

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

Solution

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

Solution

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Factor

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

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

Factor out common term

sin (x)

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

Rewrite as

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

Factor out common term

(1 + 2 cos (x))

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

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

Solving each part separately

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

1 + 2 cos (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

1 + 2 cos (x) = 0

Move

1

to the right side

1 + 2 cos (x) = 0

Subtract

1

from both sides

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

Simplify

2 cos (x) = - 1

2 cos (x) = - 1

Divide both sides by

2

2 cos (x) = - 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

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

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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

sin (x) + 1 = 0 : x = \frac{3 π}{2} + 2 πn

sin (x) + 1 = 0

Move

1

to the right side

sin (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

sin (x) = - 1

sin (x) = - 1

General solutions for

sin (x) = - 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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{3 π}{2} + 2 πn

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csc (x) = \frac{13}{5}

6csc(θ)+7=0

6 csc (θ) + 7 = 0

cos(2x)+cos(x)=0,0<= x<= 2pi

cos (2 x) + cos (x) = 0, 0 \leq x \leq 2 π

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cos (5 x) cos (3 x) - sin (5 x) sin (3 x) = \frac{3}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for sin(2x)+sin(x)+2cos(x)+1=0 ?
The general solution for sin(2x)+sin(x)+2cos(x)+1=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin,x=(3pi)/2+2pin