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

The general solution for 1+cos(θ)=-cos(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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1+cos(θ)=-cos(θ)

1 + cos (θ) = - cos (θ)

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Solution steps

1 + cos (θ) = - cos (θ)

Solve by substitution

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Combine all the solutions

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

2 cos^{2} (x) - cos (- x) - 1 = 0

3 sec^{2} (u) + 7 tan (u) = 3

3 sin (t) = 2 cos^{2} (t)

sin (θ) = \frac{12 sin ( 4 3 ^{\circ} )}{9}

2^{t a n (α)} = \frac{1}{4}

What is the general solution for 1+cos(θ)=-cos(θ) ?
The general solution for 1+cos(θ)=-cos(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin