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

The general solution for sin^2(x)-5cos(x)+5=0 is x=2pin

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

Solution

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

Solution

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

5 + sin^{2} (x) - 5 cos (x)

Use the Pythagorean identity:

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

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

= 5 + 1 - cos^{2} (x) - 5 cos (x)

Simplify

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

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

Solve by substitution

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

Let:

cos (x) = u

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

6 - u^{2} - 5 u = 0 : u = - 6, u = 1

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

Write in the standard form

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

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

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

(- 5)^{2} - 4 (- 1) \cdot 6 = 7

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

Apply rule

- (- a) = a

= (- 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

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 5 ) \pm 7}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 7}{2 ( - 1 )}, u_{2} = \frac{- ( - 5 ) - 7}{2 ( - 1 )}

u = \frac{- ( - 5 ) + 7}{2 ( - 1 )} : - 6

\frac{- ( - 5 ) + 7}{2 ( - 1 )}

Remove parentheses:

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

= \frac{5 + 7}{- 2 \cdot 1}

Add the numbers:

5 + 7 = 12

= \frac{12}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{12}{- 2}

Apply the fraction rule:

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

= - \frac{12}{2}

Divide the numbers:

\frac{12}{2} = 6

= - 6

u = \frac{- ( - 5 ) - 7}{2 ( - 1 )} : 1

\frac{- ( - 5 ) - 7}{2 ( - 1 )}

Remove parentheses:

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

= \frac{5 - 7}{- 2 \cdot 1}

Subtract the numbers:

5 - 7 = - 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 = - 6, u = 1

Substitute back

u = cos (x)

cos (x) = - 6, cos (x) = 1

cos (x) = - 6, cos (x) = 1

cos (x) = - 6 :

No Solution

cos (x) = - 6

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 1 : x = 2 πn

cos (x) = 1

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

Combine all the solutions

x = 2 πn

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Frequently Asked Questions (FAQ)

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