What is the general solution for 2sin^2(x)-5sin(x)-4=0 ?

The general solution for 2sin^2(x)-5sin(x)-4=0 is x=-0.69119…+2pin,x=pi+0.69119…+2pin

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Popular >

2sin^2(x)-5sin(x)-4=0

Solution

2 sin^{2} (x) - 5 sin (x) - 4 = 0

Solution

x = - 0.69119 \dots + 2 πn, x = π + 0.69119 \dots + 2 πn

Degrees

x = - 39.60257 \dots^{\circ} + 36 0^{\circ} n, x = 219.60257 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (x) - 5 sin (x) - 4 = 0

Solve by substitution

2 sin^{2} (x) - 5 sin (x) - 4 = 0

Let:

sin (x) = u

2 u^{2} - 5 u - 4 = 0

2 u^{2} - 5 u - 4 = 0 : u = \frac{5 + 57}{4}, u = \frac{5 - 57}{4}

2 u^{2} - 5 u - 4 = 0

Solve with the quadratic formula

2 u^{2} - 5 u - 4 = 0

Quadratic Equation Formula:

For

a = 2, b = - 5, c = - 4

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

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

(- 5)^{2} - 4 \cdot 2 (- 4) = 57

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

Apply rule

- (- a) = a

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 2 \cdot 4

Multiply the numbers:

4 \cdot 2 \cdot 4 = 32

= 5^{2} + 32

5^{2} = 25

= 25 + 32

Add the numbers:

25 + 32 = 57

= 57

u_{1, 2} = \frac{- ( - 5 ) \pm 57}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 57}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 57}{2 \cdot 2}

u = \frac{- ( - 5 ) + 57}{2 \cdot 2} : \frac{5 + 57}{4}

\frac{- ( - 5 ) + 57}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 + 57}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{5 + 57}{4}

u = \frac{- ( - 5 ) - 57}{2 \cdot 2} : \frac{5 - 57}{4}

\frac{- ( - 5 ) - 57}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 - 57}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{5 - 57}{4}

The solutions to the quadratic equation are:

u = \frac{5 + 57}{4}, u = \frac{5 - 57}{4}

Substitute back

u = sin (x)

sin (x) = \frac{5 + 57}{4}, sin (x) = \frac{5 - 57}{4}

sin (x) = \frac{5 + 57}{4}, sin (x) = \frac{5 - 57}{4}

sin (x) = \frac{5 + 57}{4} :

No Solution

sin (x) = \frac{5 + 57}{4}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{5 - 57}{4} : x = arcsin (\frac{5 - 57}{4}) + 2 πn, x = π + arcsin (- \frac{5 - 57}{4}) + 2 πn

sin (x) = \frac{5 - 57}{4}

Apply trig inverse properties

sin (x) = \frac{5 - 57}{4}

General solutions for

sin (x) = \frac{5 - 57}{4}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{5 - 57}{4}) + 2 πn, x = π + arcsin (- \frac{5 - 57}{4}) + 2 πn

x = arcsin (\frac{5 - 57}{4}) + 2 πn, x = π + arcsin (- \frac{5 - 57}{4}) + 2 πn

Combine all the solutions

x = arcsin (\frac{5 - 57}{4}) + 2 πn, x = π + arcsin (- \frac{5 - 57}{4}) + 2 πn

Show solutions in decimal form

x = - 0.69119 \dots + 2 πn, x = π + 0.69119 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)-5sin(x)-4=0 ?
The general solution for 2sin^2(x)-5sin(x)-4=0 is x=-0.69119…+2pin,x=pi+0.69119…+2pin