What is the general solution for 6cos^2(x)-7sin(x)-1=0 ?

The general solution for 6cos^2(x)-7sin(x)-1=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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6cos^2(x)-7sin(x)-1=0

Solution

6 cos^{2} (x) - 7 sin (x) - 1 = 0

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

6 cos^{2} (x) - 7 sin (x) - 1 = 0

Rewrite using trig identities

- 1 + 6 cos^{2} (x) - 7 sin (x)

Use the Pythagorean identity:

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

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

= - 1 + 6 (1 - sin^{2} (x)) - 7 sin (x)

Simplify

- 1 + 6 (1 - sin^{2} (x)) - 7 sin (x) : - 6 sin^{2} (x) - 7 sin (x) + 5

- 1 + 6 (1 - sin^{2} (x)) - 7 sin (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)

= - 1 + 6 - 6 sin^{2} (x) - 7 sin (x)

Add/Subtract the numbers:

- 1 + 6 = 5

= - 6 sin^{2} (x) - 7 sin (x) + 5

= - 6 sin^{2} (x) - 7 sin (x) + 5

5 - 6 sin^{2} (x) - 7 sin (x) = 0

Solve by substitution

5 - 6 sin^{2} (x) - 7 sin (x) = 0

Let:

sin (x) = u

5 - 6 u^{2} - 7 u = 0

5 - 6 u^{2} - 7 u = 0 : u = - \frac{5}{3}, u = \frac{1}{2}

5 - 6 u^{2} - 7 u = 0

Write in the standard form

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

- 6 u^{2} - 7 u + 5 = 0

Solve with the quadratic formula

- 6 u^{2} - 7 u + 5 = 0

Quadratic Equation Formula:

For

a = - 6, b = - 7, c = 5

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 6 ) \cdot 5}{2 ( - 6 )}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 6 ) \cdot 5}{2 ( - 6 )}

(- 7)^{2} - 4 (- 6) \cdot 5 = 13

(- 7)^{2} - 4 (- 6) \cdot 5

Apply rule

- (- a) = a

= (- 7)^{2} + 4 \cdot 6 \cdot 5

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} + 4 \cdot 6 \cdot 5

Multiply the numbers:

4 \cdot 6 \cdot 5 = 120

= 7^{2} + 120

7^{2} = 49

= 49 + 120

Add the numbers:

49 + 120 = 169

= 169

Factor the number:

169 = 1 3^{2}

= 1 3^{2}

Apply radical rule:

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- ( - 7 ) \pm 13}{2 ( - 6 )}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 13}{2 ( - 6 )}, u_{2} = \frac{- ( - 7 ) - 13}{2 ( - 6 )}

u = \frac{- ( - 7 ) + 13}{2 ( - 6 )} : - \frac{5}{3}

\frac{- ( - 7 ) + 13}{2 ( - 6 )}

Remove parentheses:

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

= \frac{7 + 13}{- 2 \cdot 6}

Add the numbers:

7 + 13 = 20

= \frac{20}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{20}{- 12}

Apply the fraction rule:

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

= - \frac{20}{12}

Cancel the common factor:

4

= - \frac{5}{3}

u = \frac{- ( - 7 ) - 13}{2 ( - 6 )} : \frac{1}{2}

\frac{- ( - 7 ) - 13}{2 ( - 6 )}

Remove parentheses:

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

= \frac{7 - 13}{- 2 \cdot 6}

Subtract the numbers:

7 - 13 = - 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{5}{3}, u = \frac{1}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{5}{3}, sin (x) = \frac{1}{2}

sin (x) = - \frac{5}{3}, sin (x) = \frac{1}{2}

sin (x) = - \frac{5}{3} :

No Solution

sin (x) = - \frac{5}{3}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

General solutions for

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

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 6cos^2(x)-7sin(x)-1=0 ?
The general solution for 6cos^2(x)-7sin(x)-1=0 is x= pi/6+2pin,x=(5pi)/6+2pin