What is the general solution for 2cos(2x)+4sin(x)+1=0 ?

The general solution for 2cos(2x)+4sin(x)+1=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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2cos(2x)+4sin(x)+1=0

Solution

2 cos (2 x) + 4 sin (x) + 1 = 0

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

2 cos (2 x) + 4 sin (x) + 1 = 0

Rewrite using trig identities

1 + 2 cos (2 x) + 4 sin (x)

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Simplify

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

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

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

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

Add the numbers:

1 + 2 = 3

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 4, b = 4, c = 3

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

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

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

4^{2} - 4 (- 4) \cdot 3

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 4 \cdot 3

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 4^{2} + 48

4^{2} = 16

= 16 + 48

Add the numbers:

16 + 48 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

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

Separate the solutions

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

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

\frac{- 4 + 8}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 4 + 8 = 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{- 4 - 8}{2 ( - 4 )} : \frac{3}{2}

\frac{- 4 - 8}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 4 - 8 = - 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

Popular Examples

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6 sin^{2} (θ) - 17 sin (θ) + 14 = - 4 sin (θ) + 9

sin(x)-2sin(x)cos(x)=0,0<= x<= 2pi

sin (x) - 2 sin (x) cos (x) = 0, 0 \leq x \leq 2 π

cos(x+pi)-cos(x)-1=0

cos (x + π) - cos (x) - 1 = 0

sin(x)= 15/17

sin (x) = \frac{15}{17}

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

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

Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2x)+4sin(x)+1=0 ?
The general solution for 2cos(2x)+4sin(x)+1=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin