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

The general solution for 4cos^2(x)-2sin(x)+1=0 is x=1.10987…+2pin,x=pi-1.10987…+2pin

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

Solution

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

Solution

x = 1.10987 \dots + 2 πn, x = π - 1.10987 \dots + 2 πn

Degrees

x = 63.59125 \dots^{\circ} + 36 0^{\circ} n, x = 116.40874 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

1 - 2 sin (x) + 4 (1 - sin^{2} (x)) : - 4 sin^{2} (x) - 2 sin (x) + 5

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

4 \cdot 1 = 4

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

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

Simplify

1 - 2 sin (x) + 4 - 4 sin^{2} (x) : - 4 sin^{2} (x) - 2 sin (x) + 5

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

Group like terms

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

Add the numbers:

1 + 4 = 5

= - 4 sin^{2} (x) - 2 sin (x) + 5

= - 4 sin^{2} (x) - 2 sin (x) + 5

= - 4 sin^{2} (x) - 2 sin (x) + 5

5 - 2 sin (x) - 4 sin^{2} (x) = 0

Solve by substitution

5 - 2 sin (x) - 4 sin^{2} (x) = 0

Let:

sin (x) = u

5 - 2 u - 4 u^{2} = 0

5 - 2 u - 4 u^{2} = 0 : u = - \frac{1 + 21}{4}, u = \frac{21 - 1}{4}

5 - 2 u - 4 u^{2} = 0

Write in the standard form

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

- 4 u^{2} - 2 u + 5 = 0

Solve with the quadratic formula

- 4 u^{2} - 2 u + 5 = 0

Quadratic Equation Formula:

For

a = - 4, b = - 2, c = 5

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

(- 2)^{2} - 4 (- 4) \cdot 5 = 221

(- 2)^{2} - 4 (- 4) \cdot 5

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 4 \cdot 5

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

= 2^{2} + 4 \cdot 4 \cdot 5

Multiply the numbers:

4 \cdot 4 \cdot 5 = 80

= 2^{2} + 80

2^{2} = 4

= 4 + 80

Add the numbers:

4 + 80 = 84

= 84

Prime factorization of

84 : 2^{2} \cdot 3 \cdot 7

84

84

divides by

284 = 42 \cdot 2

= 2 \cdot 42

42

divides by

242 = 21 \cdot 2

= 2 \cdot 2 \cdot 21

21

divides by

321 = 7 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 7

2, 3, 7

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3 \cdot 7

= 2^{2} \cdot 3 \cdot 7

= 2^{2} \cdot 3 \cdot 7

Apply radical rule:

= 2^{2} 3 \cdot 7

Apply radical rule:

2^{2} = 2

= 2 3 \cdot 7

Refine

= 221

u_{1, 2} = \frac{- ( - 2 ) \pm 2 21}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 21}{2 ( - 4 )}, u_{2} = \frac{- ( - 2 ) - 2 21}{2 ( - 4 )}

u = \frac{- ( - 2 ) + 2 21}{2 ( - 4 )} : - \frac{1 + 21}{4}

\frac{- ( - 2 ) + 2 21}{2 ( - 4 )}

Remove parentheses:

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

= \frac{2 + 2 21}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 + 2 21}{- 8}

Apply the fraction rule:

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

= - \frac{2 + 2 21}{8}

Cancel

\frac{2 + 2 21}{8} : \frac{1 + 21}{4}

\frac{2 + 2 21}{8}

Factor

2 + 221 : 2 (1 + 21)

2 + 221

Rewrite as

= 2 \cdot 1 + 221

Factor out common term

2

= 2 (1 + 21)

= \frac{2 ( 1 + 21 )}{8}

Cancel the common factor:

2

= \frac{1 + 21}{4}

= - \frac{1 + 21}{4}

u = \frac{- ( - 2 ) - 2 21}{2 ( - 4 )} : \frac{21 - 1}{4}

\frac{- ( - 2 ) - 2 21}{2 ( - 4 )}

Remove parentheses:

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

= \frac{2 - 2 21}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 - 2 21}{- 8}

Apply the fraction rule:

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

2 - 221 = - (221 - 2)

= \frac{2 21 - 2}{8}

Factor

221 - 2 : 2 (21 - 1)

221 - 2

Rewrite as

= 221 - 2 \cdot 1

Factor out common term

2

= 2 (21 - 1)

= \frac{2 ( 21 - 1 )}{8}

Cancel the common factor:

2

= \frac{21 - 1}{4}

The solutions to the quadratic equation are:

u = - \frac{1 + 21}{4}, u = \frac{21 - 1}{4}

Substitute back

u = sin (x)

sin (x) = - \frac{1 + 21}{4}, sin (x) = \frac{21 - 1}{4}

sin (x) = - \frac{1 + 21}{4}, sin (x) = \frac{21 - 1}{4}

sin (x) = - \frac{1 + 21}{4} :

No Solution

sin (x) = - \frac{1 + 21}{4}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{21 - 1}{4} : x = arcsin (\frac{21 - 1}{4}) + 2 πn, x = π - arcsin (\frac{21 - 1}{4}) + 2 πn

sin (x) = \frac{21 - 1}{4}

Apply trig inverse properties

sin (x) = \frac{21 - 1}{4}

General solutions for

sin (x) = \frac{21 - 1}{4}

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

x = arcsin (\frac{21 - 1}{4}) + 2 πn, x = π - arcsin (\frac{21 - 1}{4}) + 2 πn

x = arcsin (\frac{21 - 1}{4}) + 2 πn, x = π - arcsin (\frac{21 - 1}{4}) + 2 πn

Combine all the solutions

x = arcsin (\frac{21 - 1}{4}) + 2 πn, x = π - arcsin (\frac{21 - 1}{4}) + 2 πn

Show solutions in decimal form

x = 1.10987 \dots + 2 πn, x = π - 1.10987 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-2sin(x)+1=0 ?
The general solution for 4cos^2(x)-2sin(x)+1=0 is x=1.10987…+2pin,x=pi-1.10987…+2pin