What is the general solution for 8cos^2(θ)-3cos(θ)=0,0<= θ<= 180 ?

Q: What is the general solution for 8cos^2(θ)-3cos(θ)=0,0<= θ<= 180 ?

The general solution for 8cos^2(θ)-3cos(θ)=0,0<= θ<= 180 is θ=1.18639…,θ=90

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8cos^2(θ)-3cos(θ)=0,0<= θ<= 180

Solution

8 cos^{2} (θ) - 3 cos (θ) = 0, 0^{\circ} \leq θ \leq 18 0^{\circ}

Solution

θ = 1.18639 \dots, θ = 9 0^{\circ}

Radians

θ = 1.18639 \dots, θ = \frac{π}{2}

Solution steps

8 cos^{2} (θ) - 3 cos (θ) = 0, 0^{\circ} \leq θ \leq 18 0^{\circ}

Solve by substitution

8 cos^{2} (θ) - 3 cos (θ) = 0

Let:

cos (θ) = u

8 u^{2} - 3 u = 0

8 u^{2} - 3 u = 0 : u = \frac{3}{8}, u = 0

8 u^{2} - 3 u = 0

Solve with the quadratic formula

8 u^{2} - 3 u = 0

Quadratic Equation Formula:

For

a = 8, b = - 3, c = 0

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 8 \cdot 0}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 8 \cdot 0}{2 \cdot 8}

(- 3)^{2} - 4 \cdot 8 \cdot 0 = 3

(- 3)^{2} - 4 \cdot 8 \cdot 0

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 3^{2} - 4 \cdot 8 \cdot 0

Apply rule

0 \cdot a = 0

= 3^{2} - 0

3^{2} - 0 = 3^{2}

= 3^{2}

Apply radical rule:

n a^{n} = a,

assuming

a \geq 0

= 3

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

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 3}{2 \cdot 8}, u_{2} = \frac{- ( - 3 ) - 3}{2 \cdot 8}

u = \frac{- ( - 3 ) + 3}{2 \cdot 8} : \frac{3}{8}

\frac{- ( - 3 ) + 3}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{3 + 3}{2 \cdot 8}

Add the numbers:

3 + 3 = 6

= \frac{6}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{6}{16}

Cancel the common factor:

2

= \frac{3}{8}

u = \frac{- ( - 3 ) - 3}{2 \cdot 8} : 0

\frac{- ( - 3 ) - 3}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{3 - 3}{2 \cdot 8}

Subtract the numbers:

3 - 3 = 0

= \frac{0}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{0}{16}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = \frac{3}{8}, u = 0

Substitute back

u = cos (θ)

cos (θ) = \frac{3}{8}, cos (θ) = 0

cos (θ) = \frac{3}{8}, cos (θ) = 0

cos (θ) = \frac{3}{8}, 0 \leq θ \leq 18 0^{\circ} : θ = arccos (\frac{3}{8})

cos (θ) = \frac{3}{8}, 0 \leq θ \leq 18 0^{\circ}

Apply trig inverse properties

cos (θ) = \frac{3}{8}

General solutions for

cos (θ) = \frac{3}{8}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

θ = arccos (\frac{3}{8}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{3}{8}) + 36 0^{\circ} n

θ = arccos (\frac{3}{8}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{3}{8}) + 36 0^{\circ} n

Solutions for the range

0 \leq θ \leq 18 0^{\circ}

θ = arccos (\frac{3}{8})

cos (θ) = 0, 0 \leq θ \leq 18 0^{\circ} : θ = 9 0^{\circ}

cos (θ) = 0, 0 \leq θ \leq 18 0^{\circ}

General solutions for

cos (θ) = 0

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq θ \leq 18 0^{\circ}

θ = 9 0^{\circ}

Combine all the solutions

θ = arccos (\frac{3}{8}), θ = 9 0^{\circ}

Show solutions in decimal form

θ = 1.18639 \dots, θ = 9 0^{\circ}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 8cos^2(θ)-3cos(θ)=0,0<= θ<= 180 ?
The general solution for 8cos^2(θ)-3cos(θ)=0,0<= θ<= 180 is θ=1.18639…,θ=90