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

The general solution for 6sin^2(x)-5cos(x)-2=0 is x= pi/3+2pin,x=(5pi)/3+2pin

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

Solution

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

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

- 2 - 5 cos (x) + 6 (1 - cos^{2} (x)) : - 6 cos^{2} (x) - 5 cos (x) + 4

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

6 \cdot 1 = 6

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

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

Simplify

- 2 - 5 cos (x) + 6 - 6 cos^{2} (x) : - 6 cos^{2} (x) - 5 cos (x) + 4

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

Group like terms

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

Add/Subtract the numbers:

- 2 + 6 = 4

= - 6 cos^{2} (x) - 5 cos (x) + 4

= - 6 cos^{2} (x) - 5 cos (x) + 4

= - 6 cos^{2} (x) - 5 cos (x) + 4

4 - 5 cos (x) - 6 cos^{2} (x) = 0

Solve by substitution

4 - 5 cos (x) - 6 cos^{2} (x) = 0

Let:

cos (x) = u

4 - 5 u - 6 u^{2} = 0

4 - 5 u - 6 u^{2} = 0 : u = - \frac{4}{3}, u = \frac{1}{2}

4 - 5 u - 6 u^{2} = 0

Write in the standard form

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

- 6 u^{2} - 5 u + 4 = 0

Solve with the quadratic formula

- 6 u^{2} - 5 u + 4 = 0

Quadratic Equation Formula:

For

a = - 6, b = - 5, c = 4

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 6 \cdot 4 = 96

= 5^{2} + 96

5^{2} = 25

= 25 + 96

Add the numbers:

25 + 96 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

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

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 11}{2 ( - 6 )}, u_{2} = \frac{- ( - 5 ) - 11}{2 ( - 6 )}

u = \frac{- ( - 5 ) + 11}{2 ( - 6 )} : - \frac{4}{3}

\frac{- ( - 5 ) + 11}{2 ( - 6 )}

Remove parentheses:

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

= \frac{5 + 11}{- 2 \cdot 6}

Add the numbers:

5 + 11 = 16

= \frac{16}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{16}{- 12}

Apply the fraction rule:

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

= - \frac{16}{12}

Cancel the common factor:

4

= - \frac{4}{3}

u = \frac{- ( - 5 ) - 11}{2 ( - 6 )} : \frac{1}{2}

\frac{- ( - 5 ) - 11}{2 ( - 6 )}

Remove parentheses:

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

= \frac{5 - 11}{- 2 \cdot 6}

Subtract the numbers:

5 - 11 = - 6

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{- 12}

Apply the fraction rule:

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

= \frac{6}{12}

Cancel the common factor:

6

= \frac{1}{2}

The solutions to the quadratic equation are:

u = - \frac{4}{3}, u = \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = - \frac{4}{3}, cos (x) = \frac{1}{2}

cos (x) = - \frac{4}{3}, cos (x) = \frac{1}{2}

cos (x) = - \frac{4}{3} :

No Solution

cos (x) = - \frac{4}{3}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Popular Examples

sin(θ)+cos(θ)=sqrt(3/5)

sin (θ) + cos (θ) = \frac{3}{5}

sec(x)-4=0

sec (x) - 4 = 0

csc(θ)-cot^2(θ)+1=0

csc (θ) - cot^{2} (θ) + 1 = 0

sin(x-40)=1

sin (x - 4 0^{\circ}) = 1

sin(x)=-cos(x),-2pi<= x<= 2pi

sin (x) = - cos (x), - 2 π \leq x \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for 6sin^2(x)-5cos(x)-2=0 ?
The general solution for 6sin^2(x)-5cos(x)-2=0 is x= pi/3+2pin,x=(5pi)/3+2pin