What is the general solution for 3cos^2(θ)+7cos(θ)-8=0 ?

The general solution for 3cos^2(θ)+7cos(θ)-8=0 is θ=0.57302…+2pin,θ=2pi-0.57302…+2pin

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3cos^2(θ)+7cos(θ)-8=0

3 cos^{2} (θ) + 7 cos (θ) - 8 = 0

θ = 0.57302 \dots + 2 πn, θ = 2 π - 0.57302 \dots + 2 πn

Degrees

θ = 32.83180 \dots^{\circ} + 36 0^{\circ} n, θ = 327.16819 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 cos^{2} (θ) + 7 cos (θ) - 8 = 0

Solve by substitution

3 cos^{2} (θ) + 7 cos (θ) - 8 = 0

Let:

cos (θ) = u

3 u^{2} + 7 u - 8 = 0

3 u^{2} + 7 u - 8 = 0 : u = \frac{- 7 + 145}{6}, u = \frac{- 7 - 145}{6}

3 u^{2} + 7 u - 8 = 0

Solve with the quadratic formula

3 u^{2} + 7 u - 8 = 0

Quadratic Equation Formula:

For

a = 3, b = 7, c = - 8

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 3 ( - 8 )}{2 \cdot 3}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 3 ( - 8 )}{2 \cdot 3}

7^{2} - 4 \cdot 3 (- 8) = 145

7^{2} - 4 \cdot 3 (- 8)

Apply rule

- (- a) = a

= 7^{2} + 4 \cdot 3 \cdot 8

Multiply the numbers:

4 \cdot 3 \cdot 8 = 96

= 7^{2} + 96

7^{2} = 49

= 49 + 96

Add the numbers:

49 + 96 = 145

= 145

u_{1, 2} = \frac{- 7 \pm 145}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 7 + 145}{2 \cdot 3}, u_{2} = \frac{- 7 - 145}{2 \cdot 3}

u = \frac{- 7 + 145}{2 \cdot 3} : \frac{- 7 + 145}{6}

\frac{- 7 + 145}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 7 + 145}{6}

u = \frac{- 7 - 145}{2 \cdot 3} : \frac{- 7 - 145}{6}

\frac{- 7 - 145}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 7 - 145}{6}

The solutions to the quadratic equation are:

u = \frac{- 7 + 145}{6}, u = \frac{- 7 - 145}{6}

Substitute back

u = cos (θ)

cos (θ) = \frac{- 7 + 145}{6}, cos (θ) = \frac{- 7 - 145}{6}

cos (θ) = \frac{- 7 + 145}{6}, cos (θ) = \frac{- 7 - 145}{6}

cos (θ) = \frac{- 7 + 145}{6} : θ = arccos (\frac{- 7 + 145}{6}) + 2 πn, θ = 2 π - arccos (\frac{- 7 + 145}{6}) + 2 πn

cos (θ) = \frac{- 7 + 145}{6}

Apply trig inverse properties

cos (θ) = \frac{- 7 + 145}{6}

General solutions for

cos (θ) = \frac{- 7 + 145}{6}

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

θ = arccos (\frac{- 7 + 145}{6}) + 2 πn, θ = 2 π - arccos (\frac{- 7 + 145}{6}) + 2 πn

θ = arccos (\frac{- 7 + 145}{6}) + 2 πn, θ = 2 π - arccos (\frac{- 7 + 145}{6}) + 2 πn

cos (θ) = \frac{- 7 - 145}{6} :

No Solution

cos (θ) = \frac{- 7 - 145}{6}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = arccos (\frac{- 7 + 145}{6}) + 2 πn, θ = 2 π - arccos (\frac{- 7 + 145}{6}) + 2 πn

Show solutions in decimal form

θ = 0.57302 \dots + 2 πn, θ = 2 π - 0.57302 \dots + 2 πn

2 cos (2 a) = 2, 0 \leq θ \leq 2 π

t, y^{4} - 81 y = sin (t)

11 sin (B) - 1 = 3 sin (B) - 1

sin (5 x) - sin (3 x) - sin (x) = 0

4 arccos (2 x) = π

What is the general solution for 3cos^2(θ)+7cos(θ)-8=0 ?
The general solution for 3cos^2(θ)+7cos(θ)-8=0 is θ=0.57302…+2pin,θ=2pi-0.57302…+2pin