What is the general solution for tan(θ)= 2/(3+2cos(45)) ?

The general solution for tan(θ)= 2/(3+2cos(45)) is θ=0.42541…+180n

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tan(θ)= 2/(3+2cos(45))

tan (θ) = \frac{2}{3 + 2 cos ( 4 5 ^{\circ} )}

θ = 0.42541 \dots + 18 0^{\circ} n

Solution steps

tan (θ) = \frac{2}{3 + 2 cos ( 4 5 ^{\circ} )}

cos (4 5^{\circ}) = \frac{2}{2}

tan (θ) = \frac{2}{3 + 2 \cdot \frac{2}{2}}

Simplify

\frac{2}{3 + 2 \cdot \frac{2}{2}} : \frac{2 ( 3 - 2 )}{7}

Apply trig inverse properties

θ = arctan (\frac{2 ( 3 - 2 )}{7}) + 18 0^{\circ} n

Show solutions in decimal form

θ = 0.42541 \dots + 18 0^{\circ} n

cos (2 θ) + 6 cos^{2} (θ) = 5

sin (4 x) = - \frac{2}{2}

tan^{2} (x) - 2 tan (x) + 1 = 0

sin^{2} (x) - 2 cos^{2} (x) = 1

sin (x) = - \frac{12}{13}

What is the general solution for tan(θ)= 2/(3+2cos(45)) ?
The general solution for tan(θ)= 2/(3+2cos(45)) is θ=0.42541…+180n