Question

The sum of all values theta of belonging to [0,pi] and satisfying sin3 theta=$\sqrt{3}$/2 is

Answer 1

The sum of all values of $\theta$ is $\boxed{2\pi}$

Step-by-step explanation:

$\sin 3\theta=\frac{\sqrt{3}}{2}$

$\implies \sin 3\theta=\sin\frac{\pi}{3}$     ..... (1)

We know that if $\sin\theta=\sin\alpha$

Then the general value of $\theta$ is given by

$\theta=n\pi+(-1)^n\alpha$

Where $n=0,\pm 1, \pm 2, \pm 3, ......$

Thus, the general value of (1) is

$3\theta=n\pi+(-1)^n\frac{\pi}{3}$

$\implies \theta=\frac{n\pi}{3}+(-1)^n\frac{\pi}{9}$

Putting n = 0

$\theta=\frac{\pi}{9}$

Putting n = -1

$\theta=-\frac{\pi}{3}-\frac{\pi}{9}=-\frac{4\pi}{9}$    which does not belong in $[0, \pi]$

Putting n = 1

$\theta=\frac{\pi}{3}-\frac{\pi}{9}$

$\implies \theta=\frac{2\pi}{9}$

Putting n = -2

$\theta=-\frac{2\pi}{3}+\frac{\pi}{9}=-\frac{5\pi}{9}$    which does not belong in $[0, \pi]$

Putting n = 2

$\theta=\frac{2\pi}{3}+\frac{\pi}{9}$

$\implies \theta=\frac{7\pi}{9}$

Putting n = 3

$\theta=\frac{3\pi}{3}-\frac{\pi}{9}$

$\implies \theta=\frac{8\pi}{9}$

Putting n = 4

$\theta=\frac{4\pi}{3}+\frac{\pi}{9}$

$\implies \theta=\frac{13\pi}{9}$    which does not belong in $[0, \pi]$

Therefore, the velues of $\theta$ belonging in $[0, \pi]$ are

$\theta=\frac{\pi}{9}, \frac{2\pi}{9}, \frac{7\pi}{9}, \frac{8\pi}{9}$

The sum of the all the values is

$S=\frac{\pi}{9}+\frac{2\pi}{9}+\frac{7\pi}{9}+\frac{8\pi}{9}$

$S=\frac{18\pi}{9}$

$\implies S=2\pi$

Hope this answer is helpful.

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Answer 2

Step-by-step explanation:

The sum of all values of \thetaθ is \boxed{2\pi}

2π

Step-by-step explanation:

\sin 3\theta=\frac{\sqrt{3}}{2}sin3θ=

2

3

\implies \sin 3\theta=\sin\frac{\pi}{3}⟹sin3θ=sin

3

π

..... (1)

We know that if \sin\theta=\sin\alphasinθ=sinα

Then the general value of \thetaθ is given by

\theta=n\pi+(-1)^n\alphaθ=nπ+(−1)

n

α

Where n=0,\pm 1, \pm 2, \pm 3, ......n=0,±1,±2,±3,......

Thus, the general value of (1) is

3\theta=n\pi+(-1)^n\frac{\pi}{3}3θ=nπ+(−1)

n

3

π

\implies \theta=\frac{n\pi}{3}+(-1)^n\frac{\pi}{9}⟹θ=

3

nπ

+(−1)

n

9

π

Putting n = 0

\theta=\frac{\pi}{9}θ=

9

π

Putting n = -1

\theta=-\frac{\pi}{3}-\frac{\pi}{9}=-\frac{4\pi}{9}θ=−

3

π

−

9

π

=−

9

4π

which does not belong in [0, \pi][0,π]

Putting n = 1

\theta=\frac{\pi}{3}-\frac{\pi}{9}θ=

3

π

−

9

π

\implies \theta=\frac{2\pi}{9}⟹θ=

9

2π

Putting n = -2

\theta=-\frac{2\pi}{3}+\frac{\pi}{9}=-\frac{5\pi}{9}θ=−

3

2π

+

9

π

=−

9

5π

which does not belong in [0, \pi][0,π]

Putting n = 2

\theta=\frac{2\pi}{3}+\frac{\pi}{9}θ=

3

2π

+

9

π

\implies \theta=\frac{7\pi}{9}⟹θ=

9

7π

Putting n = 3

\theta=\frac{3\pi}{3}-\frac{\pi}{9}θ=

3

3π

−

9

π

\implies \theta=\frac{8\pi}{9}⟹θ=

9

8π

Putting n = 4

\theta=\frac{4\pi}{3}+\frac{\pi}{9}θ=

3

4π

+

9

π

\implies \theta=\frac{13\pi}{9}⟹θ=

9

13π

which does not belong in [0, \pi][0,π]

Therefore, the velues of \thetaθ belonging in [0, \pi][0,π] are

\theta=\frac{\pi}{9}, \frac{2\pi}{9}, \frac{7\pi}{9}, \frac{8\pi}{9}θ=

9

π

,

9

2π

,

9

7π

,

9

8π

The sum of the all the values is

S=\frac{\pi}{9}+\frac{2\pi}{9}+\frac{7\pi}{9}+\frac{8\pi}{9}S=

9

π

+

9

2π

+

9

7π

+

9

8π

S=\frac{18\pi}{9}S=

9

18π

\implies S=2\pi⟹S=2π

Hope this answer is helpful.

Know More:

Q: Value of theta satisfying sin7theta = sin4 theta - sin theta and theta belongs to (0, pi/2) are

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