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

The general solution for 4cos^2(x)-4sin(x)-1=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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

Solution

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

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Add/Subtract the numbers:

- 1 + 4 = 3

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

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

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

Solve by substitution

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

Let:

sin (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 = sin (x)

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

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

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

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

General solutions for

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

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

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Popular Examples

cos^2(x)+1=-2cos(x)

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

sin^3(x)-2sin(x)=0

sin^{3} (x) - 2 sin (x) = 0

2cos^3(x)=cot^3(x)

2 cos^{3} (x) = cot^{3} (x)

1/((sec^2(a)))+1/((cos^2(a)))=1

\frac{1}{( sec ^{2} ( a ) )} + \frac{1}{( cos ^{2} ( a ) )} = 1

(1-cos(a))(1+cos(a))=tan(a)sin(a)

(1 - cos (a)) (1 + cos (a)) = tan (a) sin (a)

Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-4sin(x)-1=0 ?
The general solution for 4cos^2(x)-4sin(x)-1=0 is x= pi/6+2pin,x=(5pi)/6+2pin