What is the general solution for sin(2θ)+cos(θ)=0,0<= θ<= 2pi ?

Q: What is the general solution for sin(2θ)+cos(θ)=0,0<= θ<= 2pi ?

The general solution for sin(2θ)+cos(θ)=0,0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2 ,θ=(7pi)/6 ,θ=(11pi)/6

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sin(2θ)+cos(θ)=0,0<= θ<= 2pi

Solution

sin (2 θ) + cos (θ) = 0, 0 \leq θ \leq 2 π

Solution

θ = \frac{π}{2}, θ = \frac{3 π}{2}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Degrees

θ = 9 0^{\circ}, θ = 27 0^{\circ}, θ = 21 0^{\circ}, θ = 33 0^{\circ}

Solution steps

sin (2 θ) + cos (θ) = 0, 0 \leq θ \leq 2 π

Rewrite using trig identities

cos (θ) + sin (2 θ)

Use the Double Angle identity:

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

= cos (θ) + 2 sin (θ) cos (θ)

cos (θ) + 2 cos (θ) sin (θ) = 0

Factor

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

cos (θ) + 2 cos (θ) sin (θ)

Factor out common term

cos (θ)

= cos (θ) (1 + 2 sin (θ))

cos (θ) (2 sin (θ) + 1) = 0

Solving each part separately

cos (θ) = 0 or 2 sin (θ) + 1 = 0

cos (θ) = 0, 0 \leq θ \leq 2 π : θ = \frac{π}{2}, θ = \frac{3 π}{2}

cos (θ) = 0, 0 \leq θ \leq 2 π

General solutions for

cos (θ) = 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}

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

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}

2 sin (θ) + 1 = 0, 0 \leq θ \leq 2 π : θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

2 sin (θ) + 1 = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

2 sin (θ) + 1 = 0

Subtract

1

from both sides

2 sin (θ) + 1 - 1 = 0 - 1

Simplify

2 sin (θ) = - 1

2 sin (θ) = - 1

Divide both sides by

2

2 sin (θ) = - 1

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{- 1}{2}

Simplify

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2}

General solutions for

sin (θ) = - \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}

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Combine all the solutions

θ = \frac{π}{2}, θ = \frac{3 π}{2}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Popular Examples

sin(10)=2sin(a)cos(a)

sin (1 0^{\circ}) = 2 sin (a) cos (a)

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

cos^{2} (x) = 2 + 2 sin (x), 0 \leq x \leq 2 π

cos^2(x)= 3/5

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

1-sec^2(x)=tan^2(x)

1 - sec^{2} (x) = tan^{2} (x)

csc^2(x)=(1/(cos(x)))^2

csc^{2} (x) = (\frac{1}{cos ( x )})^{2}

Frequently Asked Questions (FAQ)

What is the general solution for sin(2θ)+cos(θ)=0,0<= θ<= 2pi ?
The general solution for sin(2θ)+cos(θ)=0,0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2 ,θ=(7pi)/6 ,θ=(11pi)/6