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

The general solution for 4sin^2(x)=2cos(x)+1 is x=0.86138…+2pin,x=2pi-0.86138…+2pin

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

Solution

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

Solution

x = 0.86138 \dots + 2 πn, x = 2 π - 0.86138 \dots + 2 πn

Degrees

x = 49.35368 \dots^{\circ} + 36 0^{\circ} n, x = 310.64631 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

2 cos (x) + 1

from both sides

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

Rewrite using trig identities

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

Simplify

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 1 + 4 = 3

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

3 - 2 u - 4 u^{2} = 0 : u = - \frac{1 + 13}{4}, u = \frac{13 - 1}{4}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 4, b = - 2, c = 3

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

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

(- 2)^{2} - 4 (- 4) \cdot 3 = 213

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 4 \cdot 3

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 2^{2} + 48

2^{2} = 4

= 4 + 48

Add the numbers:

4 + 48 = 52

= 52

Prime factorization of

52 : 2^{2} \cdot 13

52

52

divides by

252 = 26 \cdot 2

= 2 \cdot 26

26

divides by

226 = 13 \cdot 2

= 2 \cdot 2 \cdot 13

2, 13

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 13

= 2^{2} \cdot 13

= 2^{2} \cdot 13

Apply radical rule:

= 13 2^{2}

Apply radical rule:

2^{2} = 2

= 213

u_{1, 2} = \frac{- ( - 2 ) \pm 2 13}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 13}{2 ( - 4 )}, u_{2} = \frac{- ( - 2 ) - 2 13}{2 ( - 4 )}

u = \frac{- ( - 2 ) + 2 13}{2 ( - 4 )} : - \frac{1 + 13}{4}

\frac{- ( - 2 ) + 2 13}{2 ( - 4 )}

Remove parentheses:

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

= \frac{2 + 2 13}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 + 2 13}{- 8}

Apply the fraction rule:

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

= - \frac{2 + 2 13}{8}

Cancel

\frac{2 + 2 13}{8} : \frac{1 + 13}{4}

\frac{2 + 2 13}{8}

Factor

2 + 213 : 2 (1 + 13)

2 + 213

Rewrite as

= 2 \cdot 1 + 213

Factor out common term

2

= 2 (1 + 13)

= \frac{2 ( 1 + 13 )}{8}

Cancel the common factor:

2

= \frac{1 + 13}{4}

= - \frac{1 + 13}{4}

u = \frac{- ( - 2 ) - 2 13}{2 ( - 4 )} : \frac{13 - 1}{4}

\frac{- ( - 2 ) - 2 13}{2 ( - 4 )}

Remove parentheses:

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

= \frac{2 - 2 13}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 - 2 13}{- 8}

Apply the fraction rule:

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

2 - 213 = - (213 - 2)

= \frac{2 13 - 2}{8}

Factor

213 - 2 : 2 (13 - 1)

213 - 2

Rewrite as

= 213 - 2 \cdot 1

Factor out common term

2

= 2 (13 - 1)

= \frac{2 ( 13 - 1 )}{8}

Cancel the common factor:

2

= \frac{13 - 1}{4}

The solutions to the quadratic equation are:

u = - \frac{1 + 13}{4}, u = \frac{13 - 1}{4}

Substitute back

u = cos (x)

cos (x) = - \frac{1 + 13}{4}, cos (x) = \frac{13 - 1}{4}

cos (x) = - \frac{1 + 13}{4}, cos (x) = \frac{13 - 1}{4}

cos (x) = - \frac{1 + 13}{4} :

No Solution

cos (x) = - \frac{1 + 13}{4}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{13 - 1}{4} : x = arccos (\frac{13 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{13 - 1}{4}) + 2 πn

cos (x) = \frac{13 - 1}{4}

Apply trig inverse properties

cos (x) = \frac{13 - 1}{4}

General solutions for

cos (x) = \frac{13 - 1}{4}

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

x = arccos (\frac{13 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{13 - 1}{4}) + 2 πn

x = arccos (\frac{13 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{13 - 1}{4}) + 2 πn

Combine all the solutions

x = arccos (\frac{13 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{13 - 1}{4}) + 2 πn

Show solutions in decimal form

x = 0.86138 \dots + 2 πn, x = 2 π - 0.86138 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)=2cos(x)+1 ?
The general solution for 4sin^2(x)=2cos(x)+1 is x=0.86138…+2pin,x=2pi-0.86138…+2pin