What is the general solution for 2cos(θ)+sqrt(2)=0,0<= θ<= 2pi ?

The general solution for 2cos(θ)+sqrt(2)=0,0<= θ<= 2pi is θ=(3pi)/4 ,θ=(5pi)/4

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2cos(θ)+sqrt(2)=0,0<= θ<= 2pi

2 cos (θ) + 2 = 0, 0 \leq θ \leq 2 π

θ = \frac{3 π}{4}, θ = \frac{5 π}{4}

Solution steps

2 cos (θ) + 2 = 0, 0 \leq θ \leq 2 π

Move

2

to the right side

2 cos (θ) = - 2

Divide both sides by

2

cos (θ) = - \frac{2}{2}

General solutions for

cos (θ) = - \frac{2}{2}

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{3 π}{4}, θ = \frac{5 π}{4}

3 tan (\frac{θ}{2}) - 1 = 0

tan (\frac{x}{2}) = 3, 0 \leq x < 2 π

sin (x) = \frac{4}{cos ( x )}

tan^{3} (x) - 2 tan (x) = tan (x)

sin^{2} (x) + 2 cos^{2} (x) = 0

What is the general solution for 2cos(θ)+sqrt(2)=0,0<= θ<= 2pi ?
The general solution for 2cos(θ)+sqrt(2)=0,0<= θ<= 2pi is θ=(3pi)/4 ,θ=(5pi)/4