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

The general solution for solvefor θ,2cos(θ)=sin(θ)+cos(θ) is θ= pi/4+pin

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solvefor θ,2cos(θ)=sin(θ)+cos(θ)

solve for

θ, 2 cos (θ) = sin (θ) + cos (θ)

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

2 cos (θ) = sin (θ) + cos (θ)

Subtract

sin (θ) + cos (θ)

from both sides

cos (θ) - sin (θ) = 0

Rewrite using trig identities

cos (θ) - sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{cos ( θ ) - sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

1 - \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

1 - tan (θ) = 0

1 - tan (θ) = 0

Move

1

to the right side

1 - tan (θ) = 0

Subtract

1

from both sides

1 - tan (θ) - 1 = 0 - 1

Simplify

- tan (θ) = - 1

- tan (θ) = - 1

Divide both sides by

- 1

- tan (θ) = - 1

Divide both sides by

- 1

\frac{- tan ( θ )}{- 1} = \frac{- 1}{- 1}

Simplify

tan (θ) = 1

tan (θ) = 1

General solutions for

tan (θ) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

x, cos (x + y) + sin (x) = 0

2 sin (x - \frac{π}{4}) + 1 = 2

3 (tan (x) - 2) = 2 tan (- 7)

6 sec (θ) + 7 = 0

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

What is the general solution for solvefor θ,2cos(θ)=sin(θ)+cos(θ) ?
The general solution for solvefor θ,2cos(θ)=sin(θ)+cos(θ) is θ= pi/4+pin