What is the general solution for 12sin^2(x)-9=0 ?

The general solution for 12sin^2(x)-9=0 is x= pi/3+2pin,x=(2pi)/3+2pin,x=(4pi)/3+2pin,x=(5pi)/3+2pin

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12sin^2(x)-9=0

Solution

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

Solution

x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

12 u^{2} - 9 = 0

12 u^{2} - 9 = 0 : u = \frac{3}{2}, u = - \frac{3}{2}

12 u^{2} - 9 = 0

Move

9

to the right side

12 u^{2} - 9 = 0

Add

9

to both sides

12 u^{2} - 9 + 9 = 0 + 9

Simplify

12 u^{2} = 9

12 u^{2} = 9

Divide both sides by

12

12 u^{2} = 9

Divide both sides by

12

\frac{12 u ^{2}}{12} = \frac{9}{12}

Simplify

u^{2} = \frac{3}{4}

u^{2} = \frac{3}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{3}{4}, u = - \frac{3}{4}

\frac{3}{4} = \frac{3}{2}

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

- \frac{3}{4} = - \frac{3}{2}

- \frac{3}{4}

Simplify

\frac{3}{4} : \frac{3}{2}

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

u = \frac{3}{2}, u = - \frac{3}{2}

Substitute back

u = sin (x)

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

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

sin (x) = \frac{3}{2} : x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

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

General solutions for

sin (x) = \frac{3}{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}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

sin (x) = - \frac{3}{2} : x = \frac{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

sin (x) = - \frac{3}{2}

General solutions for

sin (x) = - \frac{3}{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{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 12sin^2(x)-9=0 ?
The general solution for 12sin^2(x)-9=0 is x= pi/3+2pin,x=(2pi)/3+2pin,x=(4pi)/3+2pin,x=(5pi)/3+2pin