What is the general solution for sin^2(x)-9cos(x)+9=0,0<= x<= 2pi ?

Q: What is the general solution for sin^2(x)-9cos(x)+9=0,0<= x<= 2pi ?

The general solution for sin^2(x)-9cos(x)+9=0,0<= x<= 2pi is x=0,x=2pi

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Popular >

sin^2(x)-9cos(x)+9=0,0<= x<= 2pi

Solution

sin^{2} (x) - 9 cos (x) + 9 = 0, 0 \leq x \leq 2 π

Solution

x = 0, x = 2 π

Degrees

x = 0^{\circ}, x = 36 0^{\circ}

Solution steps

sin^{2} (x) - 9 cos (x) + 9 = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

9 + sin^{2} (x) - 9 cos (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 9 + 1 - cos^{2} (x) - 9 cos (x)

Simplify

= - cos^{2} (x) - 9 cos (x) + 10

10 - cos^{2} (x) - 9 cos (x) = 0

Solve by substitution

10 - cos^{2} (x) - 9 cos (x) = 0

Let:

cos (x) = u

10 - u^{2} - 9 u = 0

10 - u^{2} - 9 u = 0 : u = - 10, u = 1

10 - u^{2} - 9 u = 0

Write in the standard form

a x^{2} + b x + c = 0

- u^{2} - 9 u + 10 = 0

Solve with the quadratic formula

- u^{2} - 9 u + 10 = 0

Quadratic Equation Formula:

For

a = - 1, b = - 9, c = 10

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 ( - 1 ) \cdot 10}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 ( - 1 ) \cdot 10}{2 ( - 1 )}

(- 9)^{2} - 4 (- 1) \cdot 10 = 11

(- 9)^{2} - 4 (- 1) \cdot 10

Apply rule

- (- a) = a

= (- 9)^{2} + 4 \cdot 1 \cdot 10

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

= 9^{2} + 4 \cdot 1 \cdot 10

Multiply the numbers:

4 \cdot 1 \cdot 10 = 40

= 9^{2} + 40

9^{2} = 81

= 81 + 40

Add the numbers:

81 + 40 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

n a^{n} = a

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- ( - 9 ) \pm 11}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- ( - 9 ) + 11}{2 ( - 1 )}, u_{2} = \frac{- ( - 9 ) - 11}{2 ( - 1 )}

u = \frac{- ( - 9 ) + 11}{2 ( - 1 )} : - 10

\frac{- ( - 9 ) + 11}{2 ( - 1 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{9 + 11}{- 2 \cdot 1}

Add the numbers:

9 + 11 = 20

= \frac{20}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{20}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{20}{2}

Divide the numbers:

\frac{20}{2} = 10

= - 10

u = \frac{- ( - 9 ) - 11}{2 ( - 1 )} : 1

\frac{- ( - 9 ) - 11}{2 ( - 1 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{9 - 11}{- 2 \cdot 1}

Subtract the numbers:

9 - 11 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - 10, u = 1

Substitute back

u = cos (x)

cos (x) = - 10, cos (x) = 1

cos (x) = - 10, cos (x) = 1

cos (x) = - 10, 0 \leq x \leq 2 π :

No Solution

cos (x) = - 10, 0 \leq x \leq 2 π

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 1, 0 \leq x \leq 2 π : x = 0, x = 2 π

cos (x) = 1, 0 \leq x \leq 2 π

General solutions for

cos (x) = 1

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}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = 2 π

Combine all the solutions

x = 0, x = 2 π

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)-9cos(x)+9=0,0<= x<= 2pi ?
The general solution for sin^2(x)-9cos(x)+9=0,0<= x<= 2pi is x=0,x=2pi