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

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

The general solution for 2cos^2(θ)-3cos(θ)=0,(3pi)/2 <θ<2pi is No Solution for θ\in\mathbb{R}

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Popular >

2cos^2(θ)-3cos(θ)=0,(3pi)/2 <θ<2pi

Solution

2 cos^{2} (θ) - 3 cos (θ) = 0, \frac{3 π}{2} < θ < 2 π

Solution

No Solution for θ \in R

Solution steps

2 cos^{2} (θ) - 3 cos (θ) = 0, \frac{3 π}{2} < θ < 2 π

Solve by substitution

2 cos^{2} (θ) - 3 cos (θ) = 0

Let:

cos (θ) = u

2 u^{2} - 3 u = 0

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

2 u^{2} - 3 u = 0

Solve with the quadratic formula

2 u^{2} - 3 u = 0

Quadratic Equation Formula:

For

a = 2, b = - 3, c = 0

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

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

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

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

Apply exponent rule:

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

n

is even

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

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

Apply rule

0 \cdot a = 0

= 3^{2} - 0

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

= 3^{2}

Apply radical rule:

assuming

a \geq 0

= 3

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

3 + 3 = 6

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{6}{4}

Cancel the common factor:

2

= \frac{3}{2}

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

3 - 3 = 0

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = \frac{3}{2}, u = 0

Substitute back

u = cos (θ)

cos (θ) = \frac{3}{2}, cos (θ) = 0

cos (θ) = \frac{3}{2}, cos (θ) = 0

cos (θ) = \frac{3}{2}, \frac{3 π}{2} < θ < 2 π :

No Solution

cos (θ) = \frac{3}{2}, \frac{3 π}{2} < θ < 2 π

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = 0, \frac{3 π}{2} < θ < 2 π :

No Solution

cos (θ) = 0, \frac{3 π}{2} < θ < 2 π

General solutions for

cos (θ) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solutions for the range

\frac{3 π}{2} < θ < 2 π

No Solution

Combine all the solutions

No Solution for θ \in R

Popular Examples

cos(θ)= 9/17

cos (θ) = \frac{9}{17}

-1=(tan(x)-(3/4))/(1+tan(x)(3/4))

- 1 = \frac{tan ( x ) - ( \frac{3}{4} )}{1 + tan ( x ) ( \frac{3}{4} )}

cos(x)=-15/13

cos (x) = - \frac{15}{13}

arctan((20)/x)-arctan((400)/x)=-9.72

arctan (\frac{20}{x}) - arctan (\frac{400}{x}) = - 9.72

cos(θ)= 9/15

cos (θ) = \frac{9}{15}

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(θ)-3cos(θ)=0,(3pi)/2 <θ<2pi ?
The general solution for 2cos^2(θ)-3cos(θ)=0,(3pi)/2 <θ<2pi is No Solution for θ\in\mathbb{R}