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

The general solution for 2cos(θ)-sqrt(2)=0,0<= θ<= 2pi is θ= pi/4 ,θ=(7pi)/4

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2cos(θ)-sqrt(2)=0,0<= θ<= 2pi

2 cos (θ) - 2 = 0, 0 \leq θ \leq 2 π

θ = \frac{π}{4}, θ = \frac{7 π}{4}

Degrees

θ = 4 5^{\circ}, θ = 31 5^{\circ}

Solution steps

2 cos (θ) - 2 = 0, 0 \leq θ \leq 2 π

Move

2

to the right side

2 cos (θ) - 2 = 0

Add

2

to both sides

2 cos (θ) - 2 + 2 = 0 + 2

Simplify

2 cos (θ) = 2

2 cos (θ) = 2

Divide both sides by

2

2 cos (θ) = 2

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{2}{2}

Simplify

cos (θ) = \frac{2}{2}

cos (θ) = \frac{2}{2}

General solutions for

cos (θ) = \frac{2}{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}

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{4}, θ = \frac{7 π}{4}

5 cos^{2} (x) + 14 sin (x) cos (x) = 3 sin^{2} (x)

4 cos^{2} (x) + 4 sin^{2} (x) - 5 = 0

4 sin^{2} (x) - 7 cos (x) = 2

x, 1 sin (2 5^{\circ}) = 1.51 sin (x)

2 sin (θ) + cot (θ) = csc (θ)

What is the general solution for 2cos(θ)-sqrt(2)=0,0<= θ<= 2pi ?
The general solution for 2cos(θ)-sqrt(2)=0,0<= θ<= 2pi is θ= pi/4 ,θ=(7pi)/4