What is the general solution for cos^2(x)-5cos(x)+6=0 ?

The general solution for cos^2(x)-5cos(x)+6=0 is No Solution for x\in\mathbb{R}

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Popular >

cos^2(x)-5cos(x)+6=0

Solution

cos^{2} (x) - 5 cos (x) + 6 = 0

Solution

No Solution for x \in R

Solution steps

cos^{2} (x) - 5 cos (x) + 6 = 0

Solve by substitution

cos^{2} (x) - 5 cos (x) + 6 = 0

Let:

cos (x) = u

u^{2} - 5 u + 6 = 0

u^{2} - 5 u + 6 = 0 : u = 3, u = 2

u^{2} - 5 u + 6 = 0

Solve with the quadratic formula

u^{2} - 5 u + 6 = 0

Quadratic Equation Formula:

For

a = 1, b = - 5, c = 6

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 6 = 1

(- 5)^{2} - 4 \cdot 1 \cdot 6

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 1 \cdot 6

Multiply the numbers:

4 \cdot 1 \cdot 6 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 1}, u_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 1}

u = \frac{- ( - 5 ) + 1}{2 \cdot 1} : 3

\frac{- ( - 5 ) + 1}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{5 + 1}{2 \cdot 1}

Add the numbers:

5 + 1 = 6

= \frac{6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

u = \frac{- ( - 5 ) - 1}{2 \cdot 1} : 2

\frac{- ( - 5 ) - 1}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{5 - 1}{2 \cdot 1}

Subtract the numbers:

5 - 1 = 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

The solutions to the quadratic equation are:

u = 3, u = 2

Substitute back

u = cos (x)

cos (x) = 3, cos (x) = 2

cos (x) = 3, cos (x) = 2

cos (x) = 3 :

No Solution

cos (x) = 3

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

Popular Examples

cot^2(x/2)=3,0,0<= x<= 360

cot^{2} (\frac{x}{2}) = 3, 0, 0 \leq x \leq 360

2sin^2(x)+5sin(x)-3=0,0<= x<= 2pi

2 sin^{2} (x) + 5 sin (x) - 3 = 0, 0 \leq x \leq 2 π

sin(x)=(123sin(70))/(184.5)

sin (x) = \frac{123 sin ( 7 0 ^{\circ} )}{184.5}

cot^2(θ)=cot(θ),0<= θ<= 2pi

cot^{2} (θ) = cot (θ), 0 \leq θ \leq 2 π

cos(x)= 9/19

cos (x) = \frac{9}{19}

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-5cos(x)+6=0 ?
The general solution for cos^2(x)-5cos(x)+6=0 is No Solution for x\in\mathbb{R}