What is the general solution for sin^2(x)+2sin(x)+1=0,0<= x<2pi ?

The general solution for sin^2(x)+2sin(x)+1=0,0<= x<2pi is x=(3pi)/2

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sin^2(x)+2sin(x)+1=0,0<= x<2pi

sin^{2} (x) + 2 sin (x) + 1 = 0, 0 \leq x < 2 π

x = \frac{3 π}{2}

Degrees

x = 27 0^{\circ}

Solution steps

sin^{2} (x) + 2 sin (x) + 1 = 0, 0 \leq x < 2 π

Solve by substitution

sin^{2} (x) + 2 sin (x) + 1 = 0

Let:

sin (x) = u

u^{2} + 2 u + 1 = 0

u^{2} + 2 u + 1 = 0 : u = - 1

u^{2} + 2 u + 1 = 0

Solve with the quadratic formula

u^{2} + 2 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 2, c = 1

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

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

2^{2} - 4 \cdot 1 \cdot 1 = 0

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

u_{1, 2} = \frac{- 2 \pm 0}{2 \cdot 1}

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

u = - 1

The solution to the quadratic equation is:

u = - 1

Substitute back

u = sin (x)

sin (x) = - 1

sin (x) = - 1

sin (x) = - 1, 0 \leq x < 2 π : x = \frac{3 π}{2}

sin (x) = - 1, 0 \leq x < 2 π

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

Solutions for the range

0 \leq x < 2 π

x = \frac{3 π}{2}

Combine all the solutions

x = \frac{3 π}{2}

2 + 7 cos (θ) = 4 sin^{2} (θ)

x, sinh (x) = 0

sin (a) = 0.5

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

2 cos (3 x) = cot (3 x)

What is the general solution for sin^2(x)+2sin(x)+1=0,0<= x<2pi ?
The general solution for sin^2(x)+2sin(x)+1=0,0<= x<2pi is x=(3pi)/2