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6sin^2(x)=5+cos(x),180<= x<= 540

Solution

6 sin^{2} (x) = 5 + cos (x), 18 0^{\circ} \leq x \leq 54 0^{\circ}

Solution

x = 24 0^{\circ}, x = 48 0^{\circ}, x = 36 0^{\circ} - 1.23095 \dots, x = 1.23095 \dots + 36 0^{\circ}

Radians

x = \frac{4 π}{3}, x = \frac{8 π}{3}, x = 2 π - 1.23095 \dots, x = 1.23095 \dots + 2 π

Solution steps

6 sin^{2} (x) = 5 + cos (x), 18 0^{\circ} \leq x \leq 54 0^{\circ}

Subtract

5 + cos (x)

from both sides

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

Rewrite using trig identities

- 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)

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

Simplify

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

- 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)

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 5 + 6 = 1

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

(- 1)^{2} - 4 (- 6) \cdot 1

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 6 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 6 \cdot 1 = 24

4 \cdot 6 \cdot 1

Multiply the numbers:

4 \cdot 6 \cdot 1 = 24

= 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

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

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

\frac{- ( - 1 ) + 5}{2 ( - 6 )}

Remove parentheses:

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

= \frac{1 + 5}{- 2 \cdot 6}

Add the numbers:

1 + 5 = 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{- ( - 1 ) - 5}{2 ( - 6 )} : \frac{1}{3}

\frac{- ( - 1 ) - 5}{2 ( - 6 )}

Remove parentheses:

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

= \frac{1 - 5}{- 2 \cdot 6}

Subtract the numbers:

1 - 5 = - 4

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 4}{- 12}

Apply the fraction rule:

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

= \frac{4}{12}

Cancel the common factor:

4

= \frac{1}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = - \frac{1}{2}, 18 0^{\circ} \leq x \leq 54 0^{\circ} : x = 24 0^{\circ}, x = 48 0^{\circ}

cos (x) = - \frac{1}{2}, 18 0^{\circ} \leq x \leq 54 0^{\circ}

General solutions for

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

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Solutions for the range

18 0^{\circ} \leq x \leq 54 0^{\circ}

x = 24 0^{\circ}, x = 48 0^{\circ}

cos (x) = \frac{1}{3}, 18 0^{\circ} \leq x \leq 54 0^{\circ} : x = 36 0^{\circ} - arccos (\frac{1}{3}), x = arccos (\frac{1}{3}) + 36 0^{\circ}

cos (x) = \frac{1}{3}, 18 0^{\circ} \leq x \leq 54 0^{\circ}

Apply trig inverse properties

cos (x) = \frac{1}{3}

General solutions for

cos (x) = \frac{1}{3}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

x = arccos (\frac{1}{3}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{1}{3}) + 36 0^{\circ} n

x = arccos (\frac{1}{3}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{1}{3}) + 36 0^{\circ} n

Solutions for the range

18 0^{\circ} \leq x \leq 54 0^{\circ}

x = 36 0^{\circ} - arccos (\frac{1}{3}), x = arccos (\frac{1}{3}) + 36 0^{\circ}

Combine all the solutions

x = 24 0^{\circ}, x = 48 0^{\circ}, x = 36 0^{\circ} - arccos (\frac{1}{3}), x = arccos (\frac{1}{3}) + 36 0^{\circ}

Show solutions in decimal form

x = 24 0^{\circ}, x = 48 0^{\circ}, x = 36 0^{\circ} - 1.23095 \dots, x = 1.23095 \dots + 36 0^{\circ}

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