What is the general solution for -2cos(θ)+1=5cos(θ)+0 ?

The general solution for -2cos(θ)+1=5cos(θ)+0 is θ=1.42744…+2pin,θ=2pi-1.42744…+2pin

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-2cos(θ)+1=5cos(θ)+0

- 2 cos (θ) + 1 = 5 cos (θ) + 0

θ = 1.42744 \dots + 2 πn, θ = 2 π - 1.42744 \dots + 2 πn

Solution steps

- 2 cos (θ) + 1 = 5 cos (θ) + 0

Solve by substitution

cos (θ) = \frac{1}{7}

cos (θ) = \frac{1}{7} : θ = arccos (\frac{1}{7}) + 2 πn, θ = 2 π - arccos (\frac{1}{7}) + 2 πn

Combine all the solutions

θ = arccos (\frac{1}{7}) + 2 πn, θ = 2 π - arccos (\frac{1}{7}) + 2 πn

Show solutions in decimal form

θ = 1.42744 \dots + 2 πn, θ = 2 π - 1.42744 \dots + 2 πn

2 sin^{2} (θ) - sin (θ) = 0, 0 \leq θ < 2 π

cos (x^{2}) = \frac{1}{2}

sec^{2} (x) - 7 sec (x) = 0

e^{\frac{π}{2}} - e^{a r c t a n (1 + x^{2})} = 0

3 cot (θ) - 7 = 0

What is the general solution for -2cos(θ)+1=5cos(θ)+0 ?
The general solution for -2cos(θ)+1=5cos(θ)+0 is θ=1.42744…+2pin,θ=2pi-1.42744…+2pin