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Popular >

cos(2x)+sin(x)=4sin^2(x)

Solution

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

Solution

x = - 0.33983 \dots + 2 πn, x = π + 0.33983 \dots + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 199.47122 \dots^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

4 sin^{2} (x)

from both sides

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

Rewrite using trig identities

cos (2 x) + sin (x) - 4 sin^{2} (x)

Use the Double Angle identity:

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

= 1 - 2 sin^{2} (x) + sin (x) - 4 sin^{2} (x)

Simplify

1 - 2 sin^{2} (x) + sin (x) - 4 sin^{2} (x) : - 6 sin^{2} (x) + sin (x) + 1

1 - 2 sin^{2} (x) + sin (x) - 4 sin^{2} (x)

Group like terms

= - 2 sin^{2} (x) + sin (x) - 4 sin^{2} (x) + 1

Add similar elements:

- 2 sin^{2} (x) - 4 sin^{2} (x) = - 6 sin^{2} (x)

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 6) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 6 \cdot 1

Multiply the numbers:

4 \cdot 6 \cdot 1 = 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}{3}

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

Remove parentheses:

(- a) = - a

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

Add/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}

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

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

Remove parentheses:

(- a) = - a

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

Subtract 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}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

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 = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Show solutions in decimal form

x = - 0.33983 \dots + 2 πn, x = π + 0.33983 \dots + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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