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

The general solution for 5sin^2(x)+6cos(x)-6=0 is x=1.36943…+2pin,x=2pi-1.36943…+2pin,x=2pin

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

Solution

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

Solution

x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn, x = 2 πn

Degrees

x = 78.46304 \dots^{\circ} + 36 0^{\circ} n, x = 281.53695 \dots^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Simplify

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

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

Expand

5 (1 - cos^{2} (x)) : 5 - 5 cos^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = 1, c = cos^{2} (x)

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

Multiply the numbers:

5 \cdot 1 = 5

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

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

Add/Subtract the numbers:

- 6 + 5 = - 1

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

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

- 1 - 5 cos^{2} (x) + 6 cos (x) = 0

Solve by substitution

- 1 - 5 cos^{2} (x) + 6 cos (x) = 0

Let:

cos (x) = u

- 1 - 5 u^{2} + 6 u = 0

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

- 1 - 5 u^{2} + 6 u = 0

Write in the standard form

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

- 5 u^{2} + 6 u - 1 = 0

Solve with the quadratic formula

- 5 u^{2} + 6 u - 1 = 0

Quadratic Equation Formula:

For

a = - 5, b = 6, c = - 1

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 5 \cdot 1 = 20

= 6^{2} - 20

6^{2} = 36

= 36 - 20

Subtract the numbers:

36 - 20 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

u_{1} = \frac{- 6 + 4}{2 ( - 5 )}, u_{2} = \frac{- 6 - 4}{2 ( - 5 )}

u = \frac{- 6 + 4}{2 ( - 5 )} : \frac{1}{5}

\frac{- 6 + 4}{2 ( - 5 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 + 4}{- 2 \cdot 5}

Add/Subtract the numbers:

- 6 + 4 = - 2

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 2}{- 10}

Apply the fraction rule:

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

= \frac{2}{10}

Cancel the common factor:

2

= \frac{1}{5}

u = \frac{- 6 - 4}{2 ( - 5 )} : 1

\frac{- 6 - 4}{2 ( - 5 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 - 4}{- 2 \cdot 5}

Subtract the numbers:

- 6 - 4 = - 10

= \frac{- 10}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 10}{- 10}

Apply the fraction rule:

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

= \frac{10}{10}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = \frac{1}{5}, u = 1

Substitute back

u = cos (x)

cos (x) = \frac{1}{5}, cos (x) = 1

cos (x) = \frac{1}{5}, cos (x) = 1

cos (x) = \frac{1}{5} : x = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn

cos (x) = \frac{1}{5}

Apply trig inverse properties

cos (x) = \frac{1}{5}

General solutions for

cos (x) = \frac{1}{5}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn

x = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn

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 = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn, x = 2 πn

Show solutions in decimal form

x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn, x = 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5sin^2(x)+6cos(x)-6=0 ?
The general solution for 5sin^2(x)+6cos(x)-6=0 is x=1.36943…+2pin,x=2pi-1.36943…+2pin,x=2pin