What is the general solution for 18sin^2(x)-9sin(x)=0 ?

The general solution for 18sin^2(x)-9sin(x)=0 is x= pi/6+2pin,x=(5pi)/6+2pin,x=2pin,x=pi+2pin

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Popular >

18sin^2(x)-9sin(x)=0

Solution

18 sin^{2} (x) - 9 sin (x) = 0

Solution

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

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

18 sin^{2} (x) - 9 sin (x) = 0

Solve by substitution

18 sin^{2} (x) - 9 sin (x) = 0

Let:

sin (x) = u

18 u^{2} - 9 u = 0

18 u^{2} - 9 u = 0 : u = \frac{1}{2}, u = 0

18 u^{2} - 9 u = 0

Solve with the quadratic formula

18 u^{2} - 9 u = 0

Quadratic Equation Formula:

For

a = 18, b = - 9, c = 0

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 18 \cdot 0}{2 \cdot 18}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 18 \cdot 0}{2 \cdot 18}

(- 9)^{2} - 4 \cdot 18 \cdot 0 = 9

(- 9)^{2} - 4 \cdot 18 \cdot 0

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

= 9^{2} - 4 \cdot 18 \cdot 0

Apply rule

0 \cdot a = 0

= 9^{2} - 0

9^{2} - 0 = 9^{2}

= 9^{2}

Apply radical rule:

n a^{n} = a,

assuming

a \geq 0

= 9

u_{1, 2} = \frac{- ( - 9 ) \pm 9}{2 \cdot 18}

Separate the solutions

u_{1} = \frac{- ( - 9 ) + 9}{2 \cdot 18}, u_{2} = \frac{- ( - 9 ) - 9}{2 \cdot 18}

u = \frac{- ( - 9 ) + 9}{2 \cdot 18} : \frac{1}{2}

\frac{- ( - 9 ) + 9}{2 \cdot 18}

Apply rule

- (- a) = a

= \frac{9 + 9}{2 \cdot 18}

Add the numbers:

9 + 9 = 18

= \frac{18}{2 \cdot 18}

Multiply the numbers:

2 \cdot 18 = 36

= \frac{18}{36}

Cancel the common factor:

18

= \frac{1}{2}

u = \frac{- ( - 9 ) - 9}{2 \cdot 18} : 0

\frac{- ( - 9 ) - 9}{2 \cdot 18}

Apply rule

- (- a) = a

= \frac{9 - 9}{2 \cdot 18}

Subtract the numbers:

9 - 9 = 0

= \frac{0}{2 \cdot 18}

Multiply the numbers:

2 \cdot 18 = 36

= \frac{0}{36}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = 0

Substitute back

u = sin (x)

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

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

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Combine all the solutions

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 18sin^2(x)-9sin(x)=0 ?
The general solution for 18sin^2(x)-9sin(x)=0 is x= pi/6+2pin,x=(5pi)/6+2pin,x=2pin,x=pi+2pin