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

The general solution for 5sin(x)=2cos^2(x)-4 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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

Solution

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

Solution

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Degrees

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

2 cos^{2} (x) - 4

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = sin^{2} (x)

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Subtract the numbers:

4 - 2 = 2

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

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

2 + 2 sin^{2} (x) + 5 sin (x) = 0

Solve by substitution

2 + 2 sin^{2} (x) + 5 sin (x) = 0

Let:

sin (x) = u

2 + 2 u^{2} + 5 u = 0

2 + 2 u^{2} + 5 u = 0 : u = - \frac{1}{2}, u = - 2

2 + 2 u^{2} + 5 u = 0

Write in the standard form

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

2 u^{2} + 5 u + 2 = 0

Solve with the quadratic formula

2 u^{2} + 5 u + 2 = 0

Quadratic Equation Formula:

For

a = 2, b = 5, c = 2

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 2}{2 \cdot 2}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 2}{2 \cdot 2}

5^{2} - 4 \cdot 2 \cdot 2 = 3

5^{2} - 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

Subtract the numbers:

25 - 16 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 5 \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 5 + 3}{2 \cdot 2}, u_{2} = \frac{- 5 - 3}{2 \cdot 2}

u = \frac{- 5 + 3}{2 \cdot 2} : - \frac{1}{2}

\frac{- 5 + 3}{2 \cdot 2}

Add/Subtract the numbers:

- 5 + 3 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 5 - 3}{2 \cdot 2} : - 2

\frac{- 5 - 3}{2 \cdot 2}

Subtract the numbers:

- 5 - 3 = - 8

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 8}{4}

Apply the fraction rule:

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

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

sin (x) = - 2 :

No Solution

sin (x) = - 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Popular Examples

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2 sin (x - \frac{π}{2}) + 2 = 0

tan(2t)+tan(t)=1-tan(2t)tan(t)

tan (2 t) + tan (t) = 1 - tan (2 t) tan (t)

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-2cos(2x)=sqrt(3)

- 2 cos (2 x) = 3

(tan(x)-1)(2sin(x)-sqrt(3))=0

(tan (x) - 1) (2 sin (x) - 3) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 5sin(x)=2cos^2(x)-4 ?
The general solution for 5sin(x)=2cos^2(x)-4 is x=(7pi)/6+2pin,x=(11pi)/6+2pin