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Popular >

6sin^2(x)=6+3cos(x)

Solution

6 sin^{2} (x) = 6 + 3 cos (x)

Solution

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

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

6 sin^{2} (x) = 6 + 3 cos (x)

Subtract

6 + 3 cos (x)

from both sides

6 sin^{2} (x) - 6 - 3 cos (x) = 0

Rewrite using trig identities

- 6 - 3 cos (x) + 6 sin^{2} (x)

Use the Pythagorean identity:

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

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

= - 6 - 3 cos (x) + 6 (1 - cos^{2} (x))

Simplify

- 6 - 3 cos (x) + 6 (1 - cos^{2} (x)) : - 6 cos^{2} (x) - 3 cos (x)

- 6 - 3 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)

= - 6 - 3 cos (x) + 6 - 6 cos^{2} (x)

Simplify

- 6 - 3 cos (x) + 6 - 6 cos^{2} (x) : - 6 cos^{2} (x) - 3 cos (x)

- 6 - 3 cos (x) + 6 - 6 cos^{2} (x)

Group like terms

= - 3 cos (x) - 6 cos^{2} (x) - 6 + 6

- 6 + 6 = 0

= - 6 cos^{2} (x) - 3 cos (x)

= - 6 cos^{2} (x) - 3 cos (x)

= - 6 cos^{2} (x) - 3 cos (x)

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

Solve by substitution

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

Let:

cos (x) = u

- 3 u - 6 u^{2} = 0

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

- 3 u - 6 u^{2} = 0

Write in the standard form

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

- 6 u^{2} - 3 u = 0

Solve with the quadratic formula

- 6 u^{2} - 3 u = 0

Quadratic Equation Formula:

For

a = - 6, b = - 3, c = 0

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 6 ) \cdot 0}{2 ( - 6 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 6 ) \cdot 0}{2 ( - 6 )}

(- 3)^{2} - 4 (- 6) \cdot 0 = 3

(- 3)^{2} - 4 (- 6) \cdot 0

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 6 \cdot 0

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 6 \cdot 0

Apply rule

0 \cdot a = 0

= 3^{2} + 0

3^{2} + 0 = 3^{2}

= 3^{2}

Apply radical rule:

assuming

a \geq 0

= 3

u_{1, 2} = \frac{- ( - 3 ) \pm 3}{2 ( - 6 )}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 3}{2 ( - 6 )}, u_{2} = \frac{- ( - 3 ) - 3}{2 ( - 6 )}

u = \frac{- ( - 3 ) + 3}{2 ( - 6 )} : - \frac{1}{2}

\frac{- ( - 3 ) + 3}{2 ( - 6 )}

Remove parentheses:

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

= \frac{3 + 3}{- 2 \cdot 6}

Add the numbers:

3 + 3 = 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{- ( - 3 ) - 3}{2 ( - 6 )} : 0

\frac{- ( - 3 ) - 3}{2 ( - 6 )}

Remove parentheses:

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

= \frac{3 - 3}{- 2 \cdot 6}

Subtract the numbers:

3 - 3 = 0

= \frac{0}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{0}{- 12}

Apply the fraction rule:

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

= - \frac{0}{12}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

The solutions to the quadratic equation are:

u = - \frac{1}{2}, u = 0

Substitute back

u = cos (x)

cos (x) = - \frac{1}{2}, cos (x) = 0

cos (x) = - \frac{1}{2}, cos (x) = 0

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) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

Combine all the solutions

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

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