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

The general solution for sin^2(x)-cos(x)= 1/4 is x= pi/3+2pin,x=(5pi)/3+2pin

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

Solution

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

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

\frac{1}{4}

from both sides

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

Simplify

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

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

Convert element to fraction:

sin^{2} (x) = \frac{sin ^{2} ( x ) 4}{4}, cos (x) = \frac{cos ( x ) 4}{4}

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

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

Simplify

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 1 + 4 = 3

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 4)^{2} - 4 (- 4) \cdot 3 = 8

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 4^{2} + 48

4^{2} = 16

= 16 + 48

Add the numbers:

16 + 48 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 8}{2 ( - 4 )}, u_{2} = \frac{- ( - 4 ) - 8}{2 ( - 4 )}

u = \frac{- ( - 4 ) + 8}{2 ( - 4 )} : - \frac{3}{2}

\frac{- ( - 4 ) + 8}{2 ( - 4 )}

Remove parentheses:

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

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

Add the numbers:

4 + 8 = 12

= \frac{12}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{12}{- 8}

Apply the fraction rule:

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

= - \frac{12}{8}

Cancel the common factor:

4

= - \frac{3}{2}

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

\frac{- ( - 4 ) - 8}{2 ( - 4 )}

Remove parentheses:

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

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

Subtract the numbers:

4 - 8 = - 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

The solutions to the quadratic equation are:

u = - \frac{3}{2}, u = \frac{1}{2}

Substitute back

u = cos (x)

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

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

cos (x) = - \frac{3}{2} :

No Solution

cos (x) = - \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Popular Examples

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cos^{4} (a) = 3 + 4 cos^{2} (a) + cos^{4} (a)

cos^4(t)=1

cos^{4} (t) = 1

8cos^2(x)-12sin(x)-12=0

8 cos^{2} (x) - 12 sin (x) - 12 = 0

1-cos(x)=2

1 - cos (x) = 2

8sin^2(x)+6cos^2(x)=10

8 sin^{2} (x) + 6 cos^{2} (x) = 10

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)-cos(x)= 1/4 ?
The general solution for sin^2(x)-cos(x)= 1/4 is x= pi/3+2pin,x=(5pi)/3+2pin