What is the general solution for 4sin^2(x)+4cos(x)-5=0,0<= x<= 2pi ?

Q: What is the general solution for 4sin^2(x)+4cos(x)-5=0,0<= x<= 2pi ?

The general solution for 4sin^2(x)+4cos(x)-5=0,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3

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4sin^2(x)+4cos(x)-5=0,0<= x<= 2pi

Solution

4 sin^{2} (x) + 4 cos (x) - 5 = 0, 0 \leq x \leq 2 π

Solution

x = \frac{π}{3}, x = \frac{5 π}{3}

Degrees

x = 6 0^{\circ}, x = 30 0^{\circ}

Solution steps

4 sin^{2} (x) + 4 cos (x) - 5 = 0, 0 \leq x \leq 2 π

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

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

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

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

= \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}, 0 \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = \frac{π}{3}, x = \frac{5 π}{3}

Popular Examples

sqrt(2)sin(2θ)=1

2 sin (2 θ) = 1

cos(θ)= 1/(sqrt(2)),0<= θ<= 4pi

cos (θ) = \frac{1}{2}, 0 \leq θ \leq 4 π

sin(x)a= pi/2

sin (x) a = \frac{π}{2}

tan^2(3θ)=1

tan^{2} (3 θ) = 1

tan^2(2θ)-3=0

tan^{2} (2 θ) - 3 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)+4cos(x)-5=0,0<= x<= 2pi ?
The general solution for 4sin^2(x)+4cos(x)-5=0,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3