What is the general solution for 2sin^2(x)-15sin(x)+7=0 ?

The general solution for 2sin^2(x)-15sin(x)+7=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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2sin^2(x)-15sin(x)+7=0

Solution

2 sin^{2} (x) - 15 sin (x) + 7 = 0

Solution

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

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (x) - 15 sin (x) + 7 = 0

Solve by substitution

2 sin^{2} (x) - 15 sin (x) + 7 = 0

Let:

sin (x) = u

2 u^{2} - 15 u + 7 = 0

2 u^{2} - 15 u + 7 = 0 : u = 7, u = \frac{1}{2}

2 u^{2} - 15 u + 7 = 0

Solve with the quadratic formula

2 u^{2} - 15 u + 7 = 0

Quadratic Equation Formula:

For

a = 2, b = - 15, c = 7

u_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

(- 15)^{2} - 4 \cdot 2 \cdot 7 = 13

(- 15)^{2} - 4 \cdot 2 \cdot 7

Apply exponent rule:

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

n

is even

(- 15)^{2} = 1 5^{2}

= 1 5^{2} - 4 \cdot 2 \cdot 7

Multiply the numbers:

4 \cdot 2 \cdot 7 = 56

= 1 5^{2} - 56

1 5^{2} = 225

= 225 - 56

Subtract the numbers:

225 - 56 = 169

= 169

Factor the number:

169 = 1 3^{2}

= 1 3^{2}

Apply radical rule:

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- ( - 15 ) \pm 13}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 15 ) + 13}{2 \cdot 2}, u_{2} = \frac{- ( - 15 ) - 13}{2 \cdot 2}

u = \frac{- ( - 15 ) + 13}{2 \cdot 2} : 7

\frac{- ( - 15 ) + 13}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{15 + 13}{2 \cdot 2}

Add the numbers:

15 + 13 = 28

= \frac{28}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{28}{4}

Divide the numbers:

\frac{28}{4} = 7

= 7

u = \frac{- ( - 15 ) - 13}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 15 ) - 13}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{15 - 13}{2 \cdot 2}

Subtract the numbers:

15 - 13 = 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 = 7, u = \frac{1}{2}

Substitute back

u = sin (x)

sin (x) = 7, sin (x) = \frac{1}{2}

sin (x) = 7, sin (x) = \frac{1}{2}

sin (x) = 7 :

No Solution

sin (x) = 7

- 1 \leq sin (x) \leq 1

No Solution

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

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

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

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)-15sin(x)+7=0 ?
The general solution for 2sin^2(x)-15sin(x)+7=0 is x= pi/6+2pin,x=(5pi)/6+2pin