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Popular >

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

Solution

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

Solution

x = - 0.25268 \dots + 2 πn, x = π + 0.25268 \dots + 2 πn, x = 2 πn, x = π + 2 πn

Degrees

x = - 14.47751 \dots^{\circ} + 36 0^{\circ} n, x = 194.47751 \dots^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

- 1 + 1 = 0

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

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

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

Solve by substitution

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

Let:

sin (x) = u

- u - 4 u^{2} = 0

- u - 4 u^{2} = 0 : u = - \frac{1}{4}, u = 0

- u - 4 u^{2} = 0

Write in the standard form

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

- 4 u^{2} - u = 0

Solve with the quadratic formula

- 4 u^{2} - u = 0

Quadratic Equation Formula:

For

a = - 4, b = - 1, c = 0

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

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

(- 1)^{2} - 4 (- 4) \cdot 0 = 1

(- 1)^{2} - 4 (- 4) \cdot 0

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 4 \cdot 0

(- 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 4 \cdot 0 = 0

4 \cdot 4 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 1}{2 ( - 4 )}, u_{2} = \frac{- ( - 1 ) - 1}{2 ( - 4 )}

u = \frac{- ( - 1 ) + 1}{2 ( - 4 )} : - \frac{1}{4}

\frac{- ( - 1 ) + 1}{2 ( - 4 )}

Remove parentheses:

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

= \frac{1 + 1}{- 2 \cdot 4}

Add the numbers:

1 + 1 = 2

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2}{- 8}

Apply the fraction rule:

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

= - \frac{2}{8}

Cancel the common factor:

2

= - \frac{1}{4}

u = \frac{- ( - 1 ) - 1}{2 ( - 4 )} : 0

\frac{- ( - 1 ) - 1}{2 ( - 4 )}

Remove parentheses:

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

= \frac{1 - 1}{- 2 \cdot 4}

Subtract the numbers:

1 - 1 = 0

= \frac{0}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{0}{- 8}

Apply the fraction rule:

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

= - \frac{0}{8}

Apply rule

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

= - 0

= 0

The solutions to the quadratic equation are:

u = - \frac{1}{4}, u = 0

Substitute back

u = sin (x)

sin (x) = - \frac{1}{4}, sin (x) = 0

sin (x) = - \frac{1}{4}, sin (x) = 0

sin (x) = - \frac{1}{4} : x = arcsin (- \frac{1}{4}) + 2 πn, x = π + arcsin (\frac{1}{4}) + 2 πn

sin (x) = - \frac{1}{4}

Apply trig inverse properties

sin (x) = - \frac{1}{4}

General solutions for

sin (x) = - \frac{1}{4}

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

x = arcsin (- \frac{1}{4}) + 2 πn, x = π + arcsin (\frac{1}{4}) + 2 πn

x = arcsin (- \frac{1}{4}) + 2 πn, x = π + arcsin (\frac{1}{4}) + 2 πn

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = arcsin (- \frac{1}{4}) + 2 πn, x = π + arcsin (\frac{1}{4}) + 2 πn, x = 2 πn, x = π + 2 πn

Show solutions in decimal form

x = - 0.25268 \dots + 2 πn, x = π + 0.25268 \dots + 2 πn, x = 2 πn, x = π + 2 πn

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