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Popular >

-8cos^2(θ)+6=2

Solution

- 8 cos^{2} (θ) + 6 = 2

Solution

θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

+1

Degrees

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n, θ = - 13 5^{\circ} + 36 0^{\circ} n

Solution steps

- 8 cos^{2} (θ) + 6 = 2

Solve by substitution

- 8 cos^{2} (θ) + 6 = 2

Let:

cos (θ) = u

- 8 u^{2} + 6 = 2

- 8 u^{2} + 6 = 2 : u = \frac{1}{2}, u = - \frac{1}{2}

- 8 u^{2} + 6 = 2

Move

6

to the right side

- 8 u^{2} + 6 = 2

Subtract

6

from both sides

- 8 u^{2} + 6 - 6 = 2 - 6

Simplify

- 8 u^{2} = - 4

- 8 u^{2} = - 4

Divide both sides by

- 8

- 8 u^{2} = - 4

Divide both sides by

- 8

\frac{- 8 u ^{2}}{- 8} = \frac{- 4}{- 8}

Simplify

\frac{- 8 u ^{2}}{- 8} = \frac{- 4}{- 8}

Simplify

\frac{- 8 u ^{2}}{- 8} : u^{2}

\frac{- 8 u ^{2}}{- 8}

Apply the fraction rule:

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

= \frac{8 u ^{2}}{8}

Divide the numbers:

\frac{8}{8} = 1

= u^{2}

Simplify

\frac{- 4}{- 8} : \frac{1}{2}

\frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1}{2}, u = - \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{1}{2}, cos (θ) = - \frac{1}{2}

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

cos (θ) = \frac{1}{2}

Apply trig inverse properties

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

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

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

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

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

cos (θ) = - \frac{1}{2}

Apply trig inverse properties

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

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