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Popular >

2cos(2x)-sin(x)-1=0

Solution

2 cos (2 x) - sin (x) - 1 = 0

Solution

x = - 0.69500 \dots + 2 πn, x = π + 0.69500 \dots + 2 πn, x = 0.40105 \dots + 2 πn, x = π - 0.40105 \dots + 2 πn

Degrees

x = - 39.82077 \dots^{\circ} + 36 0^{\circ} n, x = 219.82077 \dots^{\circ} + 36 0^{\circ} n, x = 22.97865 \dots^{\circ} + 36 0^{\circ} n, x = 157.02134 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (2 x) - sin (x) - 1 = 0

Rewrite using trig identities

- 1 - sin (x) + 2 cos (2 x)

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Simplify

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

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

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 1 + 2 = 1

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

1 - u - 4 u^{2} = 0

1 - u - 4 u^{2} = 0 : u = - \frac{1 + 17}{8}, u = \frac{17 - 1}{8}

1 - u - 4 u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 1)^{2} - 4 (- 4) \cdot 1 = 17

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

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 4 \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 4 \cdot 1 = 16

4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 16

= 1 + 16

Add the numbers:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 17}{2 ( - 4 )}, u_{2} = \frac{- ( - 1 ) - 17}{2 ( - 4 )}

u = \frac{- ( - 1 ) + 17}{2 ( - 4 )} : - \frac{1 + 17}{8}

\frac{- ( - 1 ) + 17}{2 ( - 4 )}

Remove parentheses:

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

= \frac{1 + 17}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{1 + 17}{- 8}

Apply the fraction rule:

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

= - \frac{1 + 17}{8}

u = \frac{- ( - 1 ) - 17}{2 ( - 4 )} : \frac{17 - 1}{8}

\frac{- ( - 1 ) - 17}{2 ( - 4 )}

Remove parentheses:

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

= \frac{1 - 17}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{1 - 17}{- 8}

Apply the fraction rule:

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

1 - 17 = - (17 - 1)

= \frac{17 - 1}{8}

The solutions to the quadratic equation are:

u = - \frac{1 + 17}{8}, u = \frac{17 - 1}{8}

Substitute back

u = sin (x)

sin (x) = - \frac{1 + 17}{8}, sin (x) = \frac{17 - 1}{8}

sin (x) = - \frac{1 + 17}{8}, sin (x) = \frac{17 - 1}{8}

sin (x) = - \frac{1 + 17}{8} : x = arcsin (- \frac{1 + 17}{8}) + 2 πn, x = π + arcsin (\frac{1 + 17}{8}) + 2 πn

sin (x) = - \frac{1 + 17}{8}

Apply trig inverse properties

sin (x) = - \frac{1 + 17}{8}

General solutions for

sin (x) = - \frac{1 + 17}{8}

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

x = arcsin (- \frac{1 + 17}{8}) + 2 πn, x = π + arcsin (\frac{1 + 17}{8}) + 2 πn

x = arcsin (- \frac{1 + 17}{8}) + 2 πn, x = π + arcsin (\frac{1 + 17}{8}) + 2 πn

sin (x) = \frac{17 - 1}{8} : x = arcsin (\frac{17 - 1}{8}) + 2 πn, x = π - arcsin (\frac{17 - 1}{8}) + 2 πn

sin (x) = \frac{17 - 1}{8}

Apply trig inverse properties

sin (x) = \frac{17 - 1}{8}

General solutions for

sin (x) = \frac{17 - 1}{8}

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

x = arcsin (\frac{17 - 1}{8}) + 2 πn, x = π - arcsin (\frac{17 - 1}{8}) + 2 πn

x = arcsin (\frac{17 - 1}{8}) + 2 πn, x = π - arcsin (\frac{17 - 1}{8}) + 2 πn

Combine all the solutions

x = arcsin (- \frac{1 + 17}{8}) + 2 πn, x = π + arcsin (\frac{1 + 17}{8}) + 2 πn, x = arcsin (\frac{17 - 1}{8}) + 2 πn, x = π - arcsin (\frac{17 - 1}{8}) + 2 πn

Show solutions in decimal form

x = - 0.69500 \dots + 2 πn, x = π + 0.69500 \dots + 2 πn, x = 0.40105 \dots + 2 πn, x = π - 0.40105 \dots + 2 πn

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