What is the general solution for 2sin^2(x)3=7sin(x),0<= x<= 2pi ?

Q: What is the general solution for 2sin^2(x)3=7sin(x),0<= x<= 2pi ?

The general solution for 2sin^2(x)3=7sin(x),0<= x<= 2pi is x=0,x=pi,x=2pi

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2sin^2(x)3=7sin(x),0<= x<= 2pi

Solution

2 sin^{2} (x) 3 = 7 sin (x), 0 \leq x \leq 2 π

Solution

x = 0, x = π, x = 2 π

Degrees

x = 0^{\circ}, x = 18 0^{\circ}, x = 36 0^{\circ}

Solution steps

2 sin^{2} (x) \cdot 3 = 7 sin (x), 0 \leq x \leq 2 π

Solve by substitution

2 sin^{2} (x) \cdot 3 = 7 sin (x)

Let:

sin (x) = u

2 u^{2} \cdot 3 = 7 u

2 u^{2} \cdot 3 = 7 u : u = 0, u = \frac{7}{6}

2 u^{2} \cdot 3 = 7 u

Move

7 u

to the left side

2 u^{2} \cdot 3 = 7 u

Subtract

7 u

from both sides

2 u^{2} \cdot 3 - 7 u = 7 u - 7 u

Simplify

2 u^{2} \cdot 3 - 7 u = 0

2 u^{2} \cdot 3 - 7 u = 0

Factor

2 u^{2} \cdot 3 - 7 u : u (6 u - 7)

2 u^{2} \cdot 3 - 7 u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{2} = uu

= 2 \cdot 3 uu - 7 u

Factor out common term

u

= u (2 \cdot 3 u - 7)

Multiply the numbers:

2 \cdot 3 = 6

= u (6 u - 7)

u (6 u - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or 6 u - 7 = 0

Solve

6 u - 7 = 0 : u = \frac{7}{6}

6 u - 7 = 0

Move

7

to the right side

6 u - 7 = 0

Add

7

to both sides

6 u - 7 + 7 = 0 + 7

Simplify

6 u = 7

6 u = 7

Divide both sides by

6

6 u = 7

Divide both sides by

6

\frac{6 u}{6} = \frac{7}{6}

Simplify

u = \frac{7}{6}

u = \frac{7}{6}

The solutions to the quadratic equation are:

u = 0, u = \frac{7}{6}

Substitute back

u = sin (x)

sin (x) = 0, sin (x) = \frac{7}{6}

sin (x) = 0, sin (x) = \frac{7}{6}

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

sin (x) = 0, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

sin (x) = \frac{7}{6}, 0 \leq x \leq 2 π :

No Solution

sin (x) = \frac{7}{6}, 0 \leq x \leq 2 π

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = 0, x = π, x = 2 π

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solvefor L,sin(kL)=0 solve for

L, sin (k L) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)3=7sin(x),0<= x<= 2pi ?
The general solution for 2sin^2(x)3=7sin(x),0<= x<= 2pi is x=0,x=pi,x=2pi