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

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

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

Solution

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

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

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

Solution steps

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

Rewrite using trig identities

2 + cos (2 x) + sin (x)

Use the Double Angle identity:

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

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

Simplify

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = 1, c = 3

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

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

1^{2} - 4 (- 2) \cdot 3 = 5

1^{2} - 4 (- 2) \cdot 3

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 3

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 ( - 2 )}, u_{2} = \frac{- 1 - 5}{2 ( - 2 )}

u = \frac{- 1 + 5}{2 ( - 2 )} : - 1

\frac{- 1 + 5}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 2}

Add/Subtract the numbers:

- 1 + 5 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{- 4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- 1 - 5}{2 ( - 2 )} : \frac{3}{2}

\frac{- 1 - 5}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 2}

Subtract the numbers:

- 1 - 5 = - 6

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 6}{- 4}

Apply the fraction rule:

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

= \frac{6}{4}

Cancel the common factor:

2

= \frac{3}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

No Solution

sin (x) = \frac{3}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

Popular Examples

sqrt(3)tan(x)=-1

3 tan (x) = - 1

9cot^2(x)-3=0

9 cot^{2} (x) - 3 = 0

3cos(x)-4sin(x)=0

3 cos (x) - 4 sin (x) = 0

sec^2(x)-4sec(x)=0

sec^{2} (x) - 4 sec (x) = 0

sin(x)= 5/8

sin (x) = \frac{5}{8}

Frequently Asked Questions (FAQ)

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