What is the general solution for 4sin^2(x)=3,0<= x<= 180 ?

Q: What is the general solution for 4sin^2(x)=3,0<= x<= 180 ?

The general solution for 4sin^2(x)=3,0<= x<= 180 is x=60,x=120

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4sin^2(x)=3,0<= x<= 180

Solution

4 sin^{2} (x) = 3, 0 \leq x \leq 18 0^{\circ}

Solution

x = 6 0^{\circ}, x = 12 0^{\circ}

Radians

x = \frac{π}{3}, x = \frac{2 π}{3}

Solution steps

4 sin^{2} (x) = 3, 0 \leq x \leq 18 0^{\circ}

Solve by substitution

4 sin^{2} (x) = 3

Let:

sin (x) = u

4 u^{2} = 3

4 u^{2} = 3 : u = \frac{3}{2}, u = - \frac{3}{2}

4 u^{2} = 3

Divide both sides by

4

4 u^{2} = 3

Divide both sides by

4

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

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}, 0 \leq x \leq 18 0^{\circ} : x = 6 0^{\circ}, x = 12 0^{\circ}

sin (x) = \frac{3}{2}, 0 \leq x \leq 18 0^{\circ}

General solutions for

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

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

Solutions for the range

0 \leq x \leq 18 0^{\circ}

x = 6 0^{\circ}, x = 12 0^{\circ}

sin (x) = - \frac{3}{2}, 0 \leq x \leq 18 0^{\circ} :

No Solution

sin (x) = - \frac{3}{2}, 0 \leq x \leq 18 0^{\circ}

General solutions for

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

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

Solutions for the range

0 \leq x \leq 18 0^{\circ}

No Solution

Combine all the solutions

x = 6 0^{\circ}, x = 12 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)=3,0<= x<= 180 ?
The general solution for 4sin^2(x)=3,0<= x<= 180 is x=60,x=120