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1-1/2*cos(3θ)=(4-sqrt(2))/4 ,0<= θ<= 360

Solution

1 - \frac{1}{2} \cdot cos (3 θ) = \frac{4 - 2}{4}, 0^{\circ} \leq θ \leq 36 0^{\circ}

Solution

θ = 1 5^{\circ}, θ = 10 5^{\circ}, θ = 13 5^{\circ}, θ = 22 5^{\circ}, θ = 25 5^{\circ}, θ = 34 5^{\circ}

+1

Solution steps

1 - \frac{1}{2} cos (3 θ) = \frac{4 - 2}{4}, 0^{\circ} \leq θ \leq 36 0^{\circ}

Move

1

to the right side

- \frac{1}{2} cos (3 θ) = \frac{4 - 2}{4} - 1

Multiply both sides by

- 2

cos (3 θ) = - \frac{4 - 2}{2} + 2

Apply trig inverse properties

3 θ = arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n, 3 θ = 36 0^{\circ} - arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n

Solve

3 θ = arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n : θ = 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

Solve

3 θ = 36 0^{\circ} - arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n : θ = 12 0^{\circ} - 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}, θ = 12 0^{\circ} - 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

Solutions for the range

0 \leq θ \leq 36 0^{\circ}

θ = 1 5^{\circ}, θ = 10 5^{\circ}, θ = 13 5^{\circ}, θ = 22 5^{\circ}, θ = 25 5^{\circ}, θ = 34 5^{\circ}

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