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

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

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

Solution

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

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

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

Switch sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Expand

(1 - 2 u^{2}) \cdot 2 - 2 u : 2 - 4 u^{2} - 2 u

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

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

Expand

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

2 (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

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

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

Simplify

2 \cdot 1 - 2 \cdot 2 u^{2} : 2 - 4 u^{2}

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

Multiply the numbers:

2 \cdot 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= 2 - 4 u^{2}

= 2 - 4 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 4, b = - 2, c = 2

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

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

(- 2)^{2} - 4 (- 4) \cdot 2 = 6

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

Add the numbers:

4 + 32 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 6}{2 ( - 4 )}, u_{2} = \frac{- ( - 2 ) - 6}{2 ( - 4 )}

u = \frac{- ( - 2 ) + 6}{2 ( - 4 )} : - 1

\frac{- ( - 2 ) + 6}{2 ( - 4 )}

Remove parentheses:

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

= \frac{2 + 6}{- 2 \cdot 4}

Add the numbers:

2 + 6 = 8

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{- 8}

Apply the fraction rule:

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

= - \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 2 ) - 6}{2 ( - 4 )} : \frac{1}{2}

\frac{- ( - 2 ) - 6}{2 ( - 4 )}

Remove parentheses:

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

= \frac{2 - 6}{- 2 \cdot 4}

Subtract the numbers:

2 - 6 = - 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}

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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cot(2x)=1

cot (2 x) = 1

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5 tan^{2} (x) - 15 = 0

-2cos^2(x)=-cos(x)-1

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sin(θ)=-2/5

sin (θ) = - \frac{2}{5}

Frequently Asked Questions (FAQ)

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