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

Q: What is the general solution for 2sin^2(x)-5sin(x)+2=0,0<= x<= 2pi ?

The general solution for 2sin^2(x)-5sin(x)+2=0,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6

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2sin^2(x)-5sin(x)+2=0,0<= x<= 2pi

Solution

2 sin^{2} (x) - 5 sin (x) + 2 = 0, 0 \leq x \leq 2 π

Solution

x = \frac{π}{6}, x = \frac{5 π}{6}

Degrees

x = 3 0^{\circ}, x = 15 0^{\circ}

Solution steps

2 sin^{2} (x) - 5 sin (x) + 2 = 0, 0 \leq x \leq 2 π

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

Subtract the numbers:

25 - 16 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 2}

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

\frac{- ( - 5 ) + 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 + 3}{2 \cdot 2}

Add the numbers:

5 + 3 = 8

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= 2

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

\frac{- ( - 5 ) - 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 - 3}{2 \cdot 2}

Subtract the numbers:

5 - 3 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

sin (x) = 2, 0 \leq x \leq 2 π :

No Solution

sin (x) = 2, 0 \leq x \leq 2 π

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π

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{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)-5sin(x)+2=0,0<= x<= 2pi ?
The general solution for 2sin^2(x)-5sin(x)+2=0,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6