What is the general solution for -8sin(x)+8cos(2x)=0 ?

The general solution for -8sin(x)+8cos(2x)=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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-8sin(x)+8cos(2x)=0

Solution

- 8 sin (x) + 8 cos (2 x) = 0

Solution

x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

- 8 sin (x) + 8 cos (2 x) = 0

Rewrite using trig identities

8 cos (2 x) - 8 sin (x)

Use the Double Angle identity:

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

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

(1 - 2 sin^{2} (x)) \cdot 8 - 8 sin (x) = 0

Solve by substitution

(1 - 2 sin^{2} (x)) \cdot 8 - 8 sin (x) = 0

Let:

sin (x) = u

(1 - 2 u^{2}) \cdot 8 - 8 u = 0

(1 - 2 u^{2}) \cdot 8 - 8 u = 0 : u = - 1, u = \frac{1}{2}

(1 - 2 u^{2}) \cdot 8 - 8 u = 0

Expand

(1 - 2 u^{2}) \cdot 8 - 8 u : 8 - 16 u^{2} - 8 u

(1 - 2 u^{2}) \cdot 8 - 8 u

= 8 (1 - 2 u^{2}) - 8 u

Expand

8 (1 - 2 u^{2}) : 8 - 16 u^{2}

8 (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 8, b = 1, c = 2 u^{2}

= 8 \cdot 1 - 8 \cdot 2 u^{2}

Simplify

8 \cdot 1 - 8 \cdot 2 u^{2} : 8 - 16 u^{2}

8 \cdot 1 - 8 \cdot 2 u^{2}

Multiply the numbers:

8 \cdot 1 = 8

= 8 - 8 \cdot 2 u^{2}

Multiply the numbers:

8 \cdot 2 = 16

= 8 - 16 u^{2}

= 8 - 16 u^{2}

= 8 - 16 u^{2} - 8 u

8 - 16 u^{2} - 8 u = 0

Write in the standard form

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

- 16 u^{2} - 8 u + 8 = 0

Solve with the quadratic formula

- 16 u^{2} - 8 u + 8 = 0

Quadratic Equation Formula:

For

a = - 16, b = - 8, c = 8

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 8}{2 ( - 16 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 8}{2 ( - 16 )}

(- 8)^{2} - 4 (- 16) \cdot 8 = 24

(- 8)^{2} - 4 (- 16) \cdot 8

Apply rule

- (- a) = a

= (- 8)^{2} + 4 \cdot 16 \cdot 8

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 16 \cdot 8

Multiply the numbers:

4 \cdot 16 \cdot 8 = 512

= 8^{2} + 512

8^{2} = 64

= 64 + 512

Add the numbers:

64 + 512 = 576

= 576

Factor the number:

576 = 2 4^{2}

= 2 4^{2}

Apply radical rule:

2 4^{2} = 24

= 24

u_{1, 2} = \frac{- ( - 8 ) \pm 24}{2 ( - 16 )}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 24}{2 ( - 16 )}, u_{2} = \frac{- ( - 8 ) - 24}{2 ( - 16 )}

u = \frac{- ( - 8 ) + 24}{2 ( - 16 )} : - 1

\frac{- ( - 8 ) + 24}{2 ( - 16 )}

Remove parentheses:

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

= \frac{8 + 24}{- 2 \cdot 16}

Add the numbers:

8 + 24 = 32

= \frac{32}{- 2 \cdot 16}

Multiply the numbers:

2 \cdot 16 = 32

= \frac{32}{- 32}

Apply the fraction rule:

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

= - \frac{32}{32}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 8 ) - 24}{2 ( - 16 )} : \frac{1}{2}

\frac{- ( - 8 ) - 24}{2 ( - 16 )}

Remove parentheses:

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

= \frac{8 - 24}{- 2 \cdot 16}

Subtract the numbers:

8 - 24 = - 16

= \frac{- 16}{- 2 \cdot 16}

Multiply the numbers:

2 \cdot 16 = 32

= \frac{- 16}{- 32}

Apply the fraction rule:

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

= \frac{16}{32}

Cancel the common factor:

16

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

sin (x) = - 1

General solutions for

sin (x) = - 1

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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

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

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Frequently Asked Questions (FAQ)

What is the general solution for -8sin(x)+8cos(2x)=0 ?
The general solution for -8sin(x)+8cos(2x)=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin