What is the general solution for 2sin^2(x)+sin(x)-5=0 ?

The general solution for 2sin^2(x)+sin(x)-5=0 is No Solution for x\in\mathbb{R}

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2sin^2(x)+sin(x)-5=0

2 sin^{2} (x) + sin (x) - 5 = 0

No Solution for x \in R

Solution steps

2 sin^{2} (x) + sin (x) - 5 = 0

Solve by substitution

2 sin^{2} (x) + sin (x) - 5 = 0

Let:

sin (x) = u

2 u^{2} + u - 5 = 0

2 u^{2} + u - 5 = 0 : u = \frac{- 1 + 41}{4}, u = \frac{- 1 - 41}{4}

2 u^{2} + u - 5 = 0

Solve with the quadratic formula

2 u^{2} + u - 5 = 0

Quadratic Equation Formula:

For

a = 2, b = 1, c = - 5

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 5) = 41

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 5)

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 5

Multiply the numbers:

4 \cdot 2 \cdot 5 = 40

= 1 + 40

Add the numbers:

1 + 40 = 41

= 41

u_{1, 2} = \frac{- 1 \pm 41}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 1 + 41}{2 \cdot 2}, u_{2} = \frac{- 1 - 41}{2 \cdot 2}

u = \frac{- 1 + 41}{2 \cdot 2} : \frac{- 1 + 41}{4}

\frac{- 1 + 41}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 + 41}{4}

u = \frac{- 1 - 41}{2 \cdot 2} : \frac{- 1 - 41}{4}

\frac{- 1 - 41}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 - 41}{4}

The solutions to the quadratic equation are:

u = \frac{- 1 + 41}{4}, u = \frac{- 1 - 41}{4}

Substitute back

u = sin (x)

sin (x) = \frac{- 1 + 41}{4}, sin (x) = \frac{- 1 - 41}{4}

sin (x) = \frac{- 1 + 41}{4}, sin (x) = \frac{- 1 - 41}{4}

sin (x) = \frac{- 1 + 41}{4} :

No Solution

sin (x) = \frac{- 1 + 41}{4}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{- 1 - 41}{4} :

No Solution

sin (x) = \frac{- 1 - 41}{4}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

- 4 sin (x) = 4 (sin (x) - 1)

\frac{3}{tan ( x )} = (2 tan (x)) (2 cos (x))

cos (θ) = - \frac{4}{5}, sin (θ)

5 sin (4 x) = 4 cos (2 x)

arccosh (x) = 1

What is the general solution for 2sin^2(x)+sin(x)-5=0 ?
The general solution for 2sin^2(x)+sin(x)-5=0 is No Solution for x\in\mathbb{R}