What is the general solution for 4sin^2(x)=4cos(x)+5 ?

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

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4sin^2(x)=4cos(x)+5

Solution

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

Solution

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

4 cos (x) + 5

from both sides

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

Rewrite using trig identities

- 5 - 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)

= - 5 - 4 cos (x) + 4 (1 - cos^{2} (x))

Simplify

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

- 5 - 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)

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

Simplify

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

- 5 - 4 cos (x) + 4 - 4 cos^{2} (x)

Group like terms

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

Add/Subtract the numbers:

- 5 + 4 = - 1

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

- 1 - 4 u - 4 u^{2} = 0

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

- 1 - 4 u - 4 u^{2} = 0

Write in the standard form

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

- 4 u^{2} - 4 u - 1 = 0

Solve with the quadratic formula

- 4 u^{2} - 4 u - 1 = 0

Quadratic Equation Formula:

For

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

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

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

(- 4)^{2} - 4 (- 4) (- 1) = 0

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 4^{2} - 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 ( - 4 )}

u = \frac{- ( - 4 )}{2 ( - 4 )}

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

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

Remove parentheses:

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

= \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}

u = - \frac{1}{2}

The solution to the quadratic equation is:

u = - \frac{1}{2}

Substitute back

u = cos (x)

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

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

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Popular Examples

sin(3x-pi/5)=0-1

sin (3 x - \frac{π}{5}) = 0 - 1

2+sin(3θ)=3-sin(3θ)

2 + sin (3 θ) = 3 - sin (3 θ)

sin(2x)=-cos(3x)+sin(4x)

sin (2 x) = - cos (3 x) + sin (4 x)

sin(x+pi/2)=-cos(x)

sin (x + \frac{π}{2}) = - cos (x)

sin(x)=(3/5)

sin (x) = (\frac{3}{5})

Frequently Asked Questions (FAQ)

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