What is the general solution for cos(θ)=-(11)/(sqrt(170)) ?

The general solution for cos(θ)=-(11)/(sqrt(170)) is θ=2.57486…+2pin,θ=-2.57486…+2pin

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cos(θ)=-(11)/(sqrt(170))

cos (θ) = - \frac{11}{170}

θ = 2.57486 \dots + 2 πn, θ = - 2.57486 \dots + 2 πn

Radians

θ = 2.57486 \dots + 6.28318 \dots n, θ = - 2.57486 \dots + 6.28318 \dots n

Solution steps

cos (θ) = - \frac{11}{170}

Simplify

- \frac{11}{170} : - \frac{11 170}{170}

- \frac{11}{170}

Multiply by the conjugate

\frac{170}{170}

= - \frac{11 170}{170 170}

170170 = 170

170170

Apply radical rule:

a a = a

170170 = 170

= 170

= - \frac{11 170}{170}

cos (θ) = - \frac{11 170}{170}

Apply trig inverse properties

cos (θ) = - \frac{11 170}{170}

General solutions for

cos (θ) = - \frac{11 170}{170}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{11 170}{170}) + 2 πn, θ = - arccos (- \frac{11 170}{170}) + 2 πn

θ = arccos (- \frac{11 170}{170}) + 2 πn, θ = - arccos (- \frac{11 170}{170}) + 2 πn

Show solutions in decimal form

θ = 2.57486 \dots + 2 πn, θ = - 2.57486 \dots + 2 πn

2 cos^{2} (θ) - 7 cos (θ) + 3 = 0, 0 \leq θ \leq 2 π

4 sin (x) cos (x) + 2 sin (x) - 2 cos (x) = 1

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

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

2 cos^{2} (4 x) + 1 = 0

What is the general solution for cos(θ)=-(11)/(sqrt(170)) ?
The general solution for cos(θ)=-(11)/(sqrt(170)) is θ=2.57486…+2pin,θ=-2.57486…+2pin