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

The general solution for 6cos^2(x)-5sin(x)+5=0 is x= pi/2+2pin

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

Solution

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

Solution

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= 5 - 5 sin (x) + 6 (1 - sin^{2} (x))

Simplify

5 - 5 sin (x) + 6 (1 - sin^{2} (x)) : - 6 sin^{2} (x) - 5 sin (x) + 11

5 - 5 sin (x) + 6 (1 - sin^{2} (x))

Expand

6 (1 - sin^{2} (x)) : 6 - 6 sin^{2} (x)

6 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 6, b = 1, c = sin^{2} (x)

= 6 \cdot 1 - 6 sin^{2} (x)

Multiply the numbers:

6 \cdot 1 = 6

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

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

Simplify

5 - 5 sin (x) + 6 - 6 sin^{2} (x) : - 6 sin^{2} (x) - 5 sin (x) + 11

5 - 5 sin (x) + 6 - 6 sin^{2} (x)

Group like terms

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

Add the numbers:

5 + 6 = 11

= - 6 sin^{2} (x) - 5 sin (x) + 11

= - 6 sin^{2} (x) - 5 sin (x) + 11

= - 6 sin^{2} (x) - 5 sin (x) + 11

11 - 5 sin (x) - 6 sin^{2} (x) = 0

Solve by substitution

11 - 5 sin (x) - 6 sin^{2} (x) = 0

Let:

sin (x) = u

11 - 5 u - 6 u^{2} = 0

11 - 5 u - 6 u^{2} = 0 : u = - \frac{11}{6}, u = 1

11 - 5 u - 6 u^{2} = 0

Write in the standard form

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

- 6 u^{2} - 5 u + 11 = 0

Solve with the quadratic formula

- 6 u^{2} - 5 u + 11 = 0

Quadratic Equation Formula:

For

a = - 6, b = - 5, c = 11

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 6 ) \cdot 11}{2 ( - 6 )}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 6 ) \cdot 11}{2 ( - 6 )}

(- 5)^{2} - 4 (- 6) \cdot 11 = 17

(- 5)^{2} - 4 (- 6) \cdot 11

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 6 \cdot 11

Apply exponent rule:

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

n

is even

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

= 5^{2} + 4 \cdot 6 \cdot 11

Multiply the numbers:

4 \cdot 6 \cdot 11 = 264

= 5^{2} + 264

5^{2} = 25

= 25 + 264

Add the numbers:

25 + 264 = 289

= 289

Factor the number:

289 = 1 7^{2}

= 1 7^{2}

Apply radical rule:

1 7^{2} = 17

= 17

u_{1, 2} = \frac{- ( - 5 ) \pm 17}{2 ( - 6 )}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 17}{2 ( - 6 )}, u_{2} = \frac{- ( - 5 ) - 17}{2 ( - 6 )}

u = \frac{- ( - 5 ) + 17}{2 ( - 6 )} : - \frac{11}{6}

\frac{- ( - 5 ) + 17}{2 ( - 6 )}

Remove parentheses:

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

= \frac{5 + 17}{- 2 \cdot 6}

Add the numbers:

5 + 17 = 22

= \frac{22}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{22}{- 12}

Apply the fraction rule:

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

= - \frac{22}{12}

Cancel the common factor:

2

= - \frac{11}{6}

u = \frac{- ( - 5 ) - 17}{2 ( - 6 )} : 1

\frac{- ( - 5 ) - 17}{2 ( - 6 )}

Remove parentheses:

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

= \frac{5 - 17}{- 2 \cdot 6}

Subtract the numbers:

5 - 17 = - 12

= \frac{- 12}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 12}{- 12}

Apply the fraction rule:

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

= \frac{12}{12}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - \frac{11}{6}, u = 1

Substitute back

u = sin (x)

sin (x) = - \frac{11}{6}, sin (x) = 1

sin (x) = - \frac{11}{6}, sin (x) = 1

sin (x) = - \frac{11}{6} :

No Solution

sin (x) = - \frac{11}{6}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn

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2cos(x)=cos^2(x)

2 cos (x) = cos^{2} (x)

tan^2(x)-sin(x)=tan^2(x)sin^2(x)

tan^{2} (x) - sin (x) = tan^{2} (x) sin^{2} (x)

3cos(a)-1=0

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Frequently Asked Questions (FAQ)

What is the general solution for 6cos^2(x)-5sin(x)+5=0 ?
The general solution for 6cos^2(x)-5sin(x)+5=0 is x= pi/2+2pin