What is the general solution for cos(29+θ)=0.563 ?

The general solution for cos(29+θ)=0.563 is θ=0.97278…+360n-29,θ=360-0.97278…+360n-29

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cos(29+θ)=0.563

cos (2 9^{\circ} + θ) = 0.563

θ = 0.97278 \dots + 36 0^{\circ} n - 2 9^{\circ}, θ = 36 0^{\circ} - 0.97278 \dots + 36 0^{\circ} n - 2 9^{\circ}

Radians

θ = 0.97278 \dots - \frac{29 π}{180} + 2 πn, θ = 2 π - 0.97278 \dots - \frac{29 π}{180} + 2 πn

Solution steps

cos (2 9^{\circ} + θ) = 0.563

Apply trig inverse properties

cos (2 9^{\circ} + θ) = 0.563

General solutions for

cos (2 9^{\circ} + θ) = 0.563

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

2 9^{\circ} + θ = arccos (0.563) + 36 0^{\circ} n, 2 9^{\circ} + θ = 36 0^{\circ} - arccos (0.563) + 36 0^{\circ} n

2 9^{\circ} + θ = arccos (0.563) + 36 0^{\circ} n, 2 9^{\circ} + θ = 36 0^{\circ} - arccos (0.563) + 36 0^{\circ} n

Solve

2 9^{\circ} + θ = arccos (0.563) + 36 0^{\circ} n : θ = arccos (0.563) + 36 0^{\circ} n - 2 9^{\circ}

2 9^{\circ} + θ = arccos (0.563) + 36 0^{\circ} n

Move

2 9^{\circ}

to the right side

2 9^{\circ} + θ = arccos (0.563) + 36 0^{\circ} n

Subtract

2 9^{\circ}

from both sides

2 9^{\circ} + θ - 2 9^{\circ} = arccos (0.563) + 36 0^{\circ} n - 2 9^{\circ}

Simplify

θ = arccos (0.563) + 36 0^{\circ} n - 2 9^{\circ}

θ = arccos (0.563) + 36 0^{\circ} n - 2 9^{\circ}