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

The general solution for -2sin(x)-2cos(2x)=0 is x= pi/2+2pin,x=(7pi)/6+2pin,x=(11pi)/6+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

- 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}

Apply minus-plus rules

- (- a) = a

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

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

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

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

Separate the solutions

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

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

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

Apply rule

- (- 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 rule

\frac{a}{a} = 1

= 1

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

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

Apply rule

- (- 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{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{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

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

Combine all the solutions

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

Popular Examples

(cos(x))/(1+sin(x))=1-(sin(x))/(cos(x))

\frac{cos ( x )}{1 + sin ( x )} = 1 - \frac{sin ( x )}{cos ( x )}

solvefor y,2cos(5x+y)=0 solve for

y, 2 cos (5 x + y) = 0

2cos^2(x)+15sin(x)-15=0

2 cos^{2} (x) + 15 sin (x) - 15 = 0

cos(2x)=sin(x),-2pi<= x<= 2pi

cos (2 x) = sin (x), - 2 π \leq x \leq 2 π

5sec^2(θ)sin(θ)-cos(θ)=0

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

Frequently Asked Questions (FAQ)

What is the general solution for -2sin(x)-2cos(2x)=0 ?
The general solution for -2sin(x)-2cos(2x)=0 is x= pi/2+2pin,x=(7pi)/6+2pin,x=(11pi)/6+2pin