What is the general solution for cos(x)=(-1)/(sqrt(10)) ?

The general solution for cos(x)=(-1)/(sqrt(10)) is x=1.89254…+2pin,x=-1.89254…+2pin

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cos(x)=(-1)/(sqrt(10))

cos (x) = \frac{- 1}{10}

x = 1.89254 \dots + 2 πn, x = - 1.89254 \dots + 2 πn

Radians

x = 1.89254 \dots + 6.28318 \dots n, x = - 1.89254 \dots + 6.28318 \dots n

Solution steps

cos (x) = \frac{- 1}{10}

Simplify

\frac{- 1}{10} : - \frac{10}{10}

\frac{- 1}{10}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{10}

Rationalize

- \frac{1}{10} : - \frac{10}{10}

- \frac{1}{10}

Multiply by the conjugate

\frac{10}{10}

= - \frac{1 \cdot 10}{10 10}

1 \cdot 10 = 10

1010 = 10

1010

Apply radical rule:

a a = a

1010 = 10

= 10

= - \frac{10}{10}

= - \frac{10}{10}

cos (x) = - \frac{10}{10}

Apply trig inverse properties

cos (x) = - \frac{10}{10}

General solutions for

cos (x) = - \frac{10}{10}

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

x = arccos (- \frac{10}{10}) + 2 πn, x = - arccos (- \frac{10}{10}) + 2 πn

x = arccos (- \frac{10}{10}) + 2 πn, x = - arccos (- \frac{10}{10}) + 2 πn

Show solutions in decimal form

x = 1.89254 \dots + 2 πn, x = - 1.89254 \dots + 2 πn

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

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

- 1 + sin (x) = - 1

\frac{2}{2} = sin (θ)

cos (a) = 1, 8 b = 1, 5 c = 23

What is the general solution for cos(x)=(-1)/(sqrt(10)) ?
The general solution for cos(x)=(-1)/(sqrt(10)) is x=1.89254…+2pin,x=-1.89254…+2pin