What is the general solution for 2sin^2(x)-7sin(x)+3=0,0<= x<= 2pi ?

Q: What is the general solution for 2sin^2(x)-7sin(x)+3=0,0<= x<= 2pi ?

The general solution for 2sin^2(x)-7sin(x)+3=0,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6

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2sin^2(x)-7sin(x)+3=0,0<= x<= 2pi

Solution

2 sin^{2} (x) - 7 sin (x) + 3 = 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) - 7 sin (x) + 3 = 0, 0 \leq x \leq 2 π

Solve by substitution

2 sin^{2} (x) - 7 sin (x) + 3 = 0

Let:

sin (x) = u

2 u^{2} - 7 u + 3 = 0

2 u^{2} - 7 u + 3 = 0 : u = 3, u = \frac{1}{2}

2 u^{2} - 7 u + 3 = 0

Solve with the quadratic formula

2 u^{2} - 7 u + 3 = 0

Quadratic Equation Formula:

For

a = 2, b = - 7, c = 3

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 3 = 5

(- 7)^{2} - 4 \cdot 2 \cdot 3

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

Subtract the numbers:

49 - 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 2}, u_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 2}

u = \frac{- ( - 7 ) + 5}{2 \cdot 2} : 3

\frac{- ( - 7 ) + 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 + 5}{2 \cdot 2}

Add the numbers:

7 + 5 = 12

= \frac{12}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= 3

u = \frac{- ( - 7 ) - 5}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 7 ) - 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 - 5}{2 \cdot 2}

Subtract the numbers:

7 - 5 = 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 = 3, u = \frac{1}{2}

Substitute back

u = sin (x)

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

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

sin (x) = 3, 0 \leq x \leq 2 π :

No Solution

sin (x) = 3, 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}

Popular Examples

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csc (θ) = - \frac{( 2 3 )}{3}

tan(x-20)=tan(2x+10)

tan (x - 2 0^{\circ}) = tan (2 x + 10)

sin(5x)cos(x)-cos(5x)sin(x)=0

sin (5 x) cos (x) - cos (5 x) sin (x) = 0

2sin(3x+1)+1=2

2 sin (3 x + 1) + 1 = 2

2sin(3x+1)+1=1

2 sin (3 x + 1) + 1 = 1

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)-7sin(x)+3=0,0<= x<= 2pi ?
The general solution for 2sin^2(x)-7sin(x)+3=0,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6