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Popular >

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

Solution

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

Solution

x = \frac{7 π}{12} + πn, x = \frac{11 π}{12} + πn, x = \frac{0.33983 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.33983 \dots}{2} + πn

Degrees

x = 10 5^{\circ} + 18 0^{\circ} n, x = 16 5^{\circ} + 18 0^{\circ} n, x = 9.73561 \dots^{\circ} + 18 0^{\circ} n, x = 80.26438 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

sin (2 x) + 5

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

6 \cdot 1 = 6

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 5 + 6 = 1

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

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

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

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

Solve by substitution

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

Let:

sin (2 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 = sin (2 x)

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

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

sin (2 x) = - \frac{1}{2} : x = \frac{7 π}{12} + πn, x = \frac{11 π}{12} + πn

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

General solutions for

sin (2 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}

2 x = \frac{7 π}{6} + 2 πn, 2 x = \frac{11 π}{6} + 2 πn

2 x = \frac{7 π}{6} + 2 πn, 2 x = \frac{11 π}{6} + 2 πn

Solve

2 x = \frac{7 π}{6} + 2 πn : x = \frac{7 π}{12} + πn

2 x = \frac{7 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{7 π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2} : \frac{7 π}{12} + πn

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{7 π}{6}}{2} = \frac{7 π}{12}

\frac{\frac{7 π}{6}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{7 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{7 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{7 π}{12} + πn

x = \frac{7 π}{12} + πn

x = \frac{7 π}{12} + πn

x = \frac{7 π}{12} + πn

Solve

2 x = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{12} + πn

2 x = \frac{11 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{11 π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2} : \frac{11 π}{12} + πn

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{11 π}{6}}{2} = \frac{11 π}{12}

\frac{\frac{11 π}{6}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{11 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{11 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{7 π}{12} + πn, x = \frac{11 π}{12} + πn

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

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

Apply trig inverse properties

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

General solutions for

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

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

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

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

Solve

2 x = arcsin (\frac{1}{3}) + 2 πn : x = \frac{arcsin ( \frac{1}{3} )}{2} + πn

2 x = arcsin (\frac{1}{3}) + 2 πn

Divide both sides by

2

2 x = arcsin (\frac{1}{3}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{arcsin ( \frac{1}{3} )}{2} + πn

x = \frac{arcsin ( \frac{1}{3} )}{2} + πn

Solve

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

2 x = π - arcsin (\frac{1}{3}) + 2 πn

Divide both sides by

2

2 x = π - arcsin (\frac{1}{3}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

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

Combine all the solutions

x = \frac{7 π}{12} + πn, x = \frac{11 π}{12} + πn, x = \frac{arcsin ( \frac{1}{3} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

Show solutions in decimal form

x = \frac{7 π}{12} + πn, x = \frac{11 π}{12} + πn, x = \frac{0.33983 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.33983 \dots}{2} + πn

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