What is the general solution for cos(2θ)=0.4,0<= θ<= 360 ?

The general solution for cos(2θ)=0.4,0<= θ<= 360 is θ=0.57963…,θ=2.56195…,θ=3.72123…,θ=5.70354…

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cos(2θ)=0.4,0<= θ<= 360

cos (2 θ) = 0.4, 0^{\circ} \leq θ \leq 36 0^{\circ}

θ = 0.57963 \dots, θ = 2.56195 \dots, θ = 3.72123 \dots, θ = 5.70354 \dots

Solution steps

cos (2 θ) = 0.4, 0^{\circ} \leq θ \leq 36 0^{\circ}

Apply trig inverse properties

2 θ = arccos (0.4) + 36 0^{\circ} n, 2 θ = 36 0^{\circ} - arccos (0.4) + 36 0^{\circ} n

Solve

2 θ = arccos (0.4) + 36 0^{\circ} n : θ = \frac{arccos ( 0.4 )}{2} + 18 0^{\circ} n

Solve

2 θ = 36 0^{\circ} - arccos (0.4) + 36 0^{\circ} n : θ = 18 0^{\circ} - \frac{arccos ( 0.4 )}{2} + 18 0^{\circ} n

θ = \frac{arccos ( 0.4 )}{2} + 18 0^{\circ} n, θ = 18 0^{\circ} - \frac{arccos ( 0.4 )}{2} + 18 0^{\circ} n

Solutions for the range

0 \leq θ \leq 36 0^{\circ}

θ = 0.57963 \dots, θ = 2.56195 \dots, θ = 3.72123 \dots, θ = 5.70354 \dots

sin (4 x) = cos (4 x)

2 cos^{2} (x) + 9 cos (x) + 4 = 0

x, arccsc (2 x) = \frac{π}{6}

sin^{2} (x) + tan^{2} (x) = 1

cos (6 x) = sin (x - 1)

What is the general solution for cos(2θ)=0.4,0<= θ<= 360 ?
The general solution for cos(2θ)=0.4,0<= θ<= 360 is θ=0.57963…,θ=2.56195…,θ=3.72123…,θ=5.70354…