What is the general solution for 4sin^2(x)-cos(x)-1=0 ?

The general solution for 4sin^2(x)-cos(x)-1=0 is x=pi+2pin,x=0.72273…+2pin,x=2pi-0.72273…+2pin

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4sin^2(x)-cos(x)-1=0

Solution

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

Solution

x = π + 2 πn, x = 0.72273 \dots + 2 πn, x = 2 π - 0.72273 \dots + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n, x = 41.40962 \dots^{\circ} + 36 0^{\circ} n, x = 318.59037 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

- 1 - cos (x) + 4 (1 - cos^{2} (x)) : - 4 cos^{2} (x) - cos (x) + 3

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

Expand

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

4 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

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

Multiply the numbers:

4 \cdot 1 = 4

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

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

Simplify

- 1 - cos (x) + 4 - 4 cos^{2} (x) : - 4 cos^{2} (x) - cos (x) + 3

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

Group like terms

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

Add/Subtract the numbers:

- 1 + 4 = 3

= - 4 cos^{2} (x) - cos (x) + 3

= - 4 cos^{2} (x) - cos (x) + 3

= - 4 cos^{2} (x) - cos (x) + 3

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

Solve by substitution

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

Let:

cos (x) = u

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

3 - u - 4 u^{2} = 0 : u = - 1, u = \frac{3}{4}

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

Write in the standard form

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

- 4 u^{2} - u + 3 = 0

Solve with the quadratic formula

- 4 u^{2} - u + 3 = 0

Quadratic Equation Formula:

For

a = - 4, b = - 1, c = 3

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 4 ) \cdot 3}{2 ( - 4 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 4 ) \cdot 3}{2 ( - 4 )}

(- 1)^{2} - 4 (- 4) \cdot 3 = 7

(- 1)^{2} - 4 (- 4) \cdot 3

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 4 \cdot 3

(- 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 3 = 48

4 \cdot 4 \cdot 3

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 7}{2 ( - 4 )}, u_{2} = \frac{- ( - 1 ) - 7}{2 ( - 4 )}

u = \frac{- ( - 1 ) + 7}{2 ( - 4 )} : - 1

\frac{- ( - 1 ) + 7}{2 ( - 4 )}

Remove parentheses:

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

= \frac{1 + 7}{- 2 \cdot 4}

Add the numbers:

1 + 7 = 8

= \frac{8}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{- 8}

Apply the fraction rule:

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

= - \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 1 ) - 7}{2 ( - 4 )} : \frac{3}{4}

\frac{- ( - 1 ) - 7}{2 ( - 4 )}

Remove parentheses:

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

= \frac{1 - 7}{- 2 \cdot 4}

Subtract the numbers:

1 - 7 = - 6

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 6}{- 8}

Apply the fraction rule:

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

= \frac{6}{8}

Cancel the common factor:

2

= \frac{3}{4}

The solutions to the quadratic equation are:

u = - 1, u = \frac{3}{4}

Substitute back

u = cos (x)

cos (x) = - 1, cos (x) = \frac{3}{4}

cos (x) = - 1, cos (x) = \frac{3}{4}

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

cos (x) = \frac{3}{4} : x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

cos (x) = \frac{3}{4}

Apply trig inverse properties

cos (x) = \frac{3}{4}

General solutions for

cos (x) = \frac{3}{4}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

Combine all the solutions

x = π + 2 πn, x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

Show solutions in decimal form

x = π + 2 πn, x = 0.72273 \dots + 2 πn, x = 2 π - 0.72273 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)-cos(x)-1=0 ?
The general solution for 4sin^2(x)-cos(x)-1=0 is x=pi+2pin,x=0.72273…+2pin,x=2pi-0.72273…+2pin