What is the general solution for 12sin^2(θ)-3=0,0<= θ<= 2pi ?

Q: What is the general solution for 12sin^2(θ)-3=0,0<= θ<= 2pi ?

The general solution for 12sin^2(θ)-3=0,0<= θ<= 2pi is θ= pi/6 ,θ=(5pi)/6 ,θ=(7pi)/6 ,θ=(11pi)/6

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12sin^2(θ)-3=0,0<= θ<= 2pi

Solution

12 sin^{2} (θ) - 3 = 0, 0 \leq θ \leq 2 π

Solution

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Degrees

θ = 3 0^{\circ}, θ = 15 0^{\circ}, θ = 21 0^{\circ}, θ = 33 0^{\circ}

Solution steps

12 sin^{2} (θ) - 3 = 0, 0 \leq θ \leq 2 π

Solve by substitution

12 sin^{2} (θ) - 3 = 0

Let:

sin (θ) = u

12 u^{2} - 3 = 0

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

12 u^{2} - 3 = 0

Move

3

to the right side

12 u^{2} - 3 = 0

Add

3

to both sides

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

Simplify

12 u^{2} = 3

12 u^{2} = 3

Divide both sides by

12

12 u^{2} = 3

Divide both sides by

12

\frac{12 u ^{2}}{12} = \frac{3}{12}

Simplify

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{4}, u = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

Simplify

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2}

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

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{2}

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{2}

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

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

General solutions for

sin (θ) = \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}

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

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{6}, θ = \frac{5 π}{6}

sin (θ) = - \frac{1}{2}, 0 \leq θ \leq 2 π : θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

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

General solutions for

sin (θ) = - \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}

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Combine all the solutions

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Popular Examples

0=-7.5sin(pi/(10)(t-5))+8

0 = - 7.5 sin (\frac{π}{10} (t - 5)) + 8

cot(11x+1)=tan(6x+4)

cot (11 x + 1) = tan (6 x + 4)

2sin(2x)=cos(2x+30)

2 sin (2 x) = cos (2 x + 3 0^{\circ})

tan(θ)= 15/25

tan (θ) = \frac{15}{25}

tan(φ)=3

tan (φ) = 3

Frequently Asked Questions (FAQ)

What is the general solution for 12sin^2(θ)-3=0,0<= θ<= 2pi ?
The general solution for 12sin^2(θ)-3=0,0<= θ<= 2pi is θ= pi/6 ,θ=(5pi)/6 ,θ=(7pi)/6 ,θ=(11pi)/6