What is the general solution for 8sin(x)+7=4cos^2(x) ?

The general solution for 8sin(x)+7=4cos^2(x) is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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8sin(x)+7=4cos^2(x)

Solution

8 sin (x) + 7 = 4 cos^{2} (x)

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

8 sin (x) + 7 = 4 cos^{2} (x)

Subtract

4 cos^{2} (x)

from both sides

8 sin (x) + 7 - 4 cos^{2} (x) = 0

Rewrite using trig identities

7 - 4 cos^{2} (x) + 8 sin (x)

Use the Pythagorean identity:

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

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

= 7 - 4 (1 - sin^{2} (x)) + 8 sin (x)

Simplify

7 - 4 (1 - sin^{2} (x)) + 8 sin (x) : 4 sin^{2} (x) + 8 sin (x) + 3

7 - 4 (1 - sin^{2} (x)) + 8 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)

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 = 4

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

= 7 - 4 + 4 sin^{2} (x) + 8 sin (x)

Subtract the numbers:

7 - 4 = 3

= 4 sin^{2} (x) + 8 sin (x) + 3

= 4 sin^{2} (x) + 8 sin (x) + 3

3 + 4 sin^{2} (x) + 8 sin (x) = 0

Solve by substitution

3 + 4 sin^{2} (x) + 8 sin (x) = 0

Let:

sin (x) = u

3 + 4 u^{2} + 8 u = 0

3 + 4 u^{2} + 8 u = 0 : u = - \frac{1}{2}, u = - \frac{3}{2}

3 + 4 u^{2} + 8 u = 0

Write in the standard form

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

4 u^{2} + 8 u + 3 = 0

Solve with the quadratic formula

4 u^{2} + 8 u + 3 = 0

Quadratic Equation Formula:

For

a = 4, b = 8, c = 3

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 4 \cdot 3}{2 \cdot 4}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 4 \cdot 3}{2 \cdot 4}

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

8^{2} - 4 \cdot 4 \cdot 3

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 8^{2} - 48

8^{2} = 64

= 64 - 48

Subtract the numbers:

64 - 48 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

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

u = \frac{- 8 + 4}{2 \cdot 4} : - \frac{1}{2}

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

Add/Subtract the numbers:

- 8 + 4 = - 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}

u = \frac{- 8 - 4}{2 \cdot 4} : - \frac{3}{2}

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

Subtract the numbers:

- 8 - 4 = - 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}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

No Solution

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

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Frequently Asked Questions (FAQ)

What is the general solution for 8sin(x)+7=4cos^2(x) ?
The general solution for 8sin(x)+7=4cos^2(x) is x=(7pi)/6+2pin,x=(11pi)/6+2pin