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

The general solution for 2cos(θ)+sin(θ)=0 is θ=-1.10714…+pin

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2cos(θ)+sin(θ)=0

2 cos (θ) + sin (θ) = 0

θ = - 1.10714 \dots + πn

Degrees

θ = - 63.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

2 cos (θ) + sin (θ) = 0

Rewrite using trig identities

2 cos (θ) + sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

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

Simplify

2 + \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

2 + tan (θ) = 0

2 + tan (θ) = 0

Move

2

to the right side

2 + tan (θ) = 0

Subtract

2

from both sides

2 + tan (θ) - 2 = 0 - 2

Simplify

tan (θ) = - 2

tan (θ) = - 2

Apply trig inverse properties

tan (θ) = - 2

General solutions for

tan (θ) = - 2

tan (x) = - a \Rightarrow x = arctan (- a) + πn

θ = arctan (- 2) + πn

θ = arctan (- 2) + πn

Show solutions in decimal form

θ = - 1.10714 \dots + πn

2 sin^{2} (x) + sin (x) - 1 = 0, 0 < x < 2 π

5 sin (x) - 12 cos (x) = 13

sec (x) = 1.562641

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

- 4 cos (2 x) + 4 sin (x) = 0

What is the general solution for 2cos(θ)+sin(θ)=0 ?
The general solution for 2cos(θ)+sin(θ)=0 is θ=-1.10714…+pin