What is the general solution for 2cos(2θ+10)=1 ?

The general solution for 2cos(2θ+10)=1 is θ=180n+25,θ=180n+145

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2cos(2θ+10)=1

2 cos (2 θ + 1 0^{\circ}) = 1

θ = 18 0^{\circ} n + 2 5^{\circ}, θ = 18 0^{\circ} n + 14 5^{\circ}

Solution steps

2 cos (2 θ + 1 0^{\circ}) = 1

Divide both sides by

2

cos (2 θ + 1 0^{\circ}) = \frac{1}{2}

General solutions for

cos (2 θ + 1 0^{\circ}) = \frac{1}{2}

2 θ + 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n, 2 θ + 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n

Solve

2 θ + 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n : θ = 18 0^{\circ} n + 2 5^{\circ}

Solve

2 θ + 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n : θ = 18 0^{\circ} n + 14 5^{\circ}

θ = 18 0^{\circ} n + 2 5^{\circ}, θ = 18 0^{\circ} n + 14 5^{\circ}

cot (β) = \frac{- 3}{3}

2 sin^{2} (3 x) - 3 cos (3 x) + 1 = 0

sin (θ) = \frac{15}{23}

3 sin^{2} (x) + 6 sin (x) - 11 = 7 sin (x) - 9

sin^{2} (a) = 2 sin (a)

What is the general solution for 2cos(2θ+10)=1 ?
The general solution for 2cos(2θ+10)=1 is θ=180n+25,θ=180n+145