What is the general solution for cos(2θ)= 3/5 ,90<= θ<= 180 ?

The general solution for cos(2θ)= 3/5 ,90<= θ<= 180 is θ=(360-0.92729…)/2

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cos(2θ)= 3/5 ,90<= θ<= 180

cos (2 θ) = \frac{3}{5}, 9 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = \frac{36 0 ^{\circ} - 0.92729 \dots}{2}

Solution steps

cos (2 θ) = \frac{3}{5}, 9 0^{\circ} \leq θ \leq 18 0^{\circ}

Apply trig inverse properties

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

Solve

2 θ = arccos (\frac{3}{5}) + 36 0^{\circ} n : θ = \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

Solve

2 θ = 36 0^{\circ} - arccos (\frac{3}{5}) + 36 0^{\circ} n : θ = 18 0^{\circ} - \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

θ = \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n, θ = 18 0^{\circ} - \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

Solutions for the range

9 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = \frac{36 0 ^{\circ} - arccos ( \frac{3}{5} )}{2}

Show solutions in decimal form

θ = \frac{36 0 ^{\circ} - 0.92729 \dots}{2}

tan (θ) = \frac{1}{- 1}

sin^{2} (x) - 4 cos (x) - 4 = 0, 0 \leq x \leq 2 π

cos^{2} (θ) = 2 - 2 sin (θ)

x, sin (x) = 0.5

x, z = arcsin (\frac{y}{x})

What is the general solution for cos(2θ)= 3/5 ,90<= θ<= 180 ?
The general solution for cos(2θ)= 3/5 ,90<= θ<= 180 is θ=(360-0.92729…)/2