What is the general solution for cos^2(x)-6cos(x)=-9 ?

The general solution for cos^2(x)-6cos(x)=-9 is No Solution for x\in\mathbb{R}

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Popular >

cos^2(x)-6cos(x)=-9

Solution

cos^{2} (x) - 6 cos (x) = - 9

Solution

No Solution for x \in R

Solution steps

cos^{2} (x) - 6 cos (x) = - 9

Solve by substitution

cos^{2} (x) - 6 cos (x) = - 9

Let:

cos (x) = u

u^{2} - 6 u = - 9

u^{2} - 6 u = - 9 : u = 3

u^{2} - 6 u = - 9

Move

9

to the left side

u^{2} - 6 u = - 9

Add

9

to both sides

u^{2} - 6 u + 9 = - 9 + 9

Simplify

u^{2} - 6 u + 9 = 0

u^{2} - 6 u + 9 = 0

Solve with the quadratic formula

u^{2} - 6 u + 9 = 0

Quadratic Equation Formula:

For

a = 1, b = - 6, c = 9

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 9 = 0

(- 6)^{2} - 4 \cdot 1 \cdot 9

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 1 \cdot 9

Multiply the numbers:

4 \cdot 1 \cdot 9 = 36

= 6^{2} - 36

6^{2} = 36

= 36 - 36

Subtract the numbers:

36 - 36 = 0

= 0

u_{1, 2} = \frac{- ( - 6 ) \pm 0}{2 \cdot 1}

u = \frac{- ( - 6 )}{2 \cdot 1}

\frac{- ( - 6 )}{2 \cdot 1} = 3

\frac{- ( - 6 )}{2 \cdot 1}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

u = 3

The solution to the quadratic equation is:

u = 3

Substitute back

u = cos (x)

cos (x) = 3

cos (x) = 3

cos (x) = 3 :

No Solution

cos (x) = 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

Graph

Popular Examples

cos(x)-3=0

cos (x) - 3 = 0

sec(x)-tan(x)=3

sec (x) - tan (x) = 3

sqrt(2)cos(x)=0

2 cos (x) = 0

tan^2(6x)=3

tan^{2} (6 x) = 3

tan(2x)=3sin(2x)

tan (2 x) = 3 sin (2 x)

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-6cos(x)=-9 ?
The general solution for cos^2(x)-6cos(x)=-9 is No Solution for x\in\mathbb{R}