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

The general solution for sin^2(x)-6cos(x)+6=0 is x=2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = 2 πn

Degrees

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

Solution steps

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

Rewrite using trig identities

6 + sin^{2} (x) - 6 cos (x)

Use the Pythagorean identity:

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

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

= 6 + 1 - cos^{2} (x) - 6 cos (x)

Simplify

= - cos^{2} (x) - 6 cos (x) + 7

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

Solve by substitution

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

Let:

cos (x) = u

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

7 - u^{2} - 6 u = 0 : u = - 7, u = 1

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

Write in the standard form

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

- u^{2} - 6 u + 7 = 0

Solve with the quadratic formula

- u^{2} - 6 u + 7 = 0

Quadratic Equation Formula:

For

a = - 1, b = - 6, c = 7

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 1 ) \cdot 7}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 1 ) \cdot 7}{2 ( - 1 )}

(- 6)^{2} - 4 (- 1) \cdot 7 = 8

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 6^{2} + 4 \cdot 1 \cdot 7

Multiply the numbers:

4 \cdot 1 \cdot 7 = 28

= 6^{2} + 28

6^{2} = 36

= 36 + 28

Add the numbers:

36 + 28 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 6 ) \pm 8}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 8}{2 ( - 1 )}, u_{2} = \frac{- ( - 6 ) - 8}{2 ( - 1 )}

u = \frac{- ( - 6 ) + 8}{2 ( - 1 )} : - 7

\frac{- ( - 6 ) + 8}{2 ( - 1 )}

Remove parentheses:

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

= \frac{6 + 8}{- 2 \cdot 1}

Add the numbers:

6 + 8 = 14

= \frac{14}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{14}{- 2}

Apply the fraction rule:

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

= - \frac{14}{2}

Divide the numbers:

\frac{14}{2} = 7

= - 7

u = \frac{- ( - 6 ) - 8}{2 ( - 1 )} : 1

\frac{- ( - 6 ) - 8}{2 ( - 1 )}

Remove parentheses:

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

= \frac{6 - 8}{- 2 \cdot 1}

Subtract the numbers:

6 - 8 = - 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 = - 7, u = 1

Substitute back

u = cos (x)

cos (x) = - 7, cos (x) = 1

cos (x) = - 7, cos (x) = 1

cos (x) = - 7 :

No Solution

cos (x) = - 7

- 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

Popular Examples

cos(x)=(14.3)/(19)

cos (x) = \frac{14.3}{19}

-12sin(x)=0

- 12 sin (x) = 0

3cot^2(x)-5=-4

3 cot^{2} (x) - 5 = - 4

25cot^2(x)-16=0

25 cot^{2} (x) - 16 = 0

-9cos^2(θ)+9=16sin(θ)-7

- 9 cos^{2} (θ) + 9 = 16 sin (θ) - 7

Frequently Asked Questions (FAQ)

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