What is the general solution for cos^2(θ)=2cos(θ) ?

The general solution for cos^2(θ)=2cos(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin

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cos^2(θ)=2cos(θ)

Solution

cos^{2} (θ) = 2 cos (θ)

Solution

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

Degrees

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

Solution steps

cos^{2} (θ) = 2 cos (θ)

Solve by substitution

cos^{2} (θ) = 2 cos (θ)

Let:

cos (θ) = u

u^{2} = 2 u

u^{2} = 2 u : u = 2, u = 0

u^{2} = 2 u

Move

2 u

to the left side

u^{2} = 2 u

Subtract

2 u

from both sides

u^{2} - 2 u = 2 u - 2 u

Simplify

u^{2} - 2 u = 0

u^{2} - 2 u = 0

Solve with the quadratic formula

u^{2} - 2 u = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = 0

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

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

(- 2)^{2} - 4 \cdot 1 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 1 \cdot 0

Apply exponent rule:

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

n

is even

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

= 2^{2} - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

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

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

u = \frac{- ( - 2 ) + 2}{2 \cdot 1} : 2

\frac{- ( - 2 ) + 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 + 2}{2 \cdot 1}

Add the numbers:

2 + 2 = 4

= \frac{4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

u = \frac{- ( - 2 ) - 2}{2 \cdot 1} : 0

\frac{- ( - 2 ) - 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 - 2}{2 \cdot 1}

Subtract the numbers:

2 - 2 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = 2, u = 0

Substitute back

u = cos (θ)

cos (θ) = 2, cos (θ) = 0

cos (θ) = 2, cos (θ) = 0

cos (θ) = 2 :

No Solution

cos (θ) = 2

- 1 \leq cos (x) \leq 1

No Solution

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

cos (θ) = 0

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

Combine all the solutions

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

Popular Examples

2sin(x)-sin(2x)=sin(2x)

2 sin (x) - sin (2 x) = sin (2 x)

2cos^2(x)+cos(2x)-2=0

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

3cos(x)+5=7,0<= x<= 360

3 cos (x) + 5 = 7, 0^{\circ} \leq x \leq 36 0^{\circ}

cot(3x-6)=(sqrt(3))/3

cot (3 x - 6^{\circ}) = \frac{3}{3}

sec(x)=sqrt(4)

sec (x) = 4

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(θ)=2cos(θ) ?
The general solution for cos^2(θ)=2cos(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin