What is the general solution for cos(2θ)=((sqrt(57))/(11)) ?

The general solution for cos(2θ)=((sqrt(57))/(11)) is θ=(0.81433…)/2+pin,θ=pi-(0.81433…)/2+pin

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cos(2θ)=((sqrt(57))/(11))

cos (2 θ) = (\frac{57}{11})

θ = \frac{0.81433 \dots}{2} + πn, θ = π - \frac{0.81433 \dots}{2} + πn

Solution steps

cos (2 θ) = (\frac{57}{11})

Apply trig inverse properties

2 θ = arccos (\frac{57}{11}) + 2 πn, 2 θ = 2 π - arccos (\frac{57}{11}) + 2 πn

Solve

2 θ = arccos (\frac{57}{11}) + 2 πn : θ = \frac{arccos ( \frac{57}{11} )}{2} + πn

Solve

2 θ = 2 π - arccos (\frac{57}{11}) + 2 πn : θ = π - \frac{arccos ( \frac{57}{11} )}{2} + πn

θ = \frac{arccos ( \frac{57}{11} )}{2} + πn, θ = π - \frac{arccos ( \frac{57}{11} )}{2} + πn

Show solutions in decimal form

θ = \frac{0.81433 \dots}{2} + πn, θ = π - \frac{0.81433 \dots}{2} + πn

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

θ, \frac{cos ( θ )}{sin ( θ )} = - \frac{π}{6}

tan (θ) = \frac{10}{- 5.32}

2 sin (2 x) + cos (2 x) = 4 cos (2 x)

sec (t) = \frac{1}{\frac{2}{2}}

What is the general solution for cos(2θ)=((sqrt(57))/(11)) ?
The general solution for cos(2θ)=((sqrt(57))/(11)) is θ=(0.81433…)/2+pin,θ=pi-(0.81433…)/2+pin