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=(2pi)/3+2pin,x=(4pi)/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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 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)) : 5 cos (x) - 6 cos^{2} (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) : 5 cos (x) - 6 cos^{2} (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

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

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

= 5 cos (x) - 6 cos^{2} (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{1}{2}, u = \frac{4}{3}

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

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{1}{2}

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

Remove parentheses:

(- a) = - a

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

Add/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}

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

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

Remove parentheses:

(- a) = - a

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

Subtract 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}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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\frac{sin ^{22} ( a )}{sin ^{2} ( a )} = 4 - 4 sin^{2} (a)

tan^3(x)=2

tan^{3} (x) = 2

sin^3(x)=3sin(x)

sin^{3} (x) = 3 sin (x)

cos^4(x)+2sin^2(x)+6cos^2(x)+5=0

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

1+sin(2a)=sin^2(a)

1 + sin (2 a) = sin^{2} (a)

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=(2pi)/3+2pin,x=(4pi)/3+2pin