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Question: If \[1 + \sin \theta + {\sin ^2}\theta + ..... = 4 + 2\sqrt 3 \], \[0 < \theta < \pi,\theta \ne \dfr...

If 1+sin⁡θ+sin⁡2θ+.....=4+231 + \sin \theta + {\sin ^2}\theta + ..... = 4 + 2\sqrt 3 , 0<θ<π,θ≠π20 < \theta < \pi,\theta \ne \dfrac{\pi }{2} then
A. θ=π3\theta = \dfrac{\pi }{3}
B. θ=π6\theta = \dfrac{\pi }{6}
C. θ=π3orπ6\theta = \dfrac{\pi }{3}or\dfrac{\pi }{6}
D. θ=π3or2π3\theta = \dfrac{\pi }{3}or\dfrac{{2\pi }}{3}

Explanation

Solution

When we observe the given series we come to know it is related to one trigonometric function that is sine. But it is along with a combination of geometric progression with the common ratio as sin⁡θ\sin \theta . We will use the formula of G.P. to find the sum of the series on LHS and then the actual sum is already given. We will equate them to solving it and we will get the values of θ\theta . Then we will decide the answer.

Complete step by step answer:
Given the series is 1+sin⁡θ+sin⁡2θ+.....=4+231 + \sin \theta + {\sin ^2}\theta + ..... = 4 + 2\sqrt 3
We can observe that, it is a G.P. with sin⁡θ\sin \theta as the common ratio.
We know that the sum of terms in a G.P. is given by,
S=a1−rS = \dfrac{a}{{1 - r}}
Here a is the first term equals to 1 and r is the common ratio equals to sin⁡θ\sin \theta . Putting these values,
S=11−sin⁡θS = \dfrac{1}{{1 - \sin \theta }}
But the sum is already given to us, then we will equate them as;
11−sin⁡θ=4+23\dfrac{1}{{1 - \sin \theta }} = 4 + 2\sqrt 3
On transposing the terms,
1−sin⁡θ=14+231 - \sin \theta = \dfrac{1}{{4 + 2\sqrt 3 }}
Multiplying and dividing the RHS by 4−234 - 2\sqrt 3 ,
1−sin⁡θ=14+23×4−234−231 - \sin \theta = \dfrac{1}{{4 + 2\sqrt 3 }} \times \dfrac{{4 - 2\sqrt 3 }}{{4 - 2\sqrt 3 }}
The denominator now becomes,
1−sin⁡θ=4−2342−(23)21 - \sin \theta = \dfrac{{4 - 2\sqrt 3 }}{{{4^2} - {{\left( {2\sqrt 3 } \right)}^2}}}
Taking the squares,
1−sin⁡θ=4−2316−121 - \sin \theta = \dfrac{{4 - 2\sqrt 3 }}{{16 - 12}}
On subtracting we get,
1−sin⁡θ=4−2341 - \sin \theta = \dfrac{{4 - 2\sqrt 3 }}{4}
Separating the terms,
1−sin⁡θ=1−321 - \sin \theta = 1 - \dfrac{{\sqrt 3 }}{2}
Cancelling 1 from both sides,
sin⁡θ=32\sin \theta = \dfrac{{\sqrt 3 }}{2}
To find the value of θ\theta
θ=sin⁡−1(32)\theta = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)
We know that, sin⁡nπ3=32\sin \dfrac{{n\pi }}{3} = \dfrac{{\sqrt 3 }}{2}
∴θ=π3 or 2π3\therefore \theta = \dfrac{\pi }{3}\,or\,\dfrac{{2\pi }}{3} this is because 0<θ<π,θ≠π20 < \theta < \pi ,\theta \ne \dfrac{\pi }{2}

Thus option D is the correct answer.

Note: Here we can get confused about the options. But we should check the limit or restrictions on the θ\theta . It is greater than 0 but less than π\pi . Talking about 2π3\dfrac{{2\pi }}{3} it lies below or under the limit so assigned. So choose options wisely.
Second thing that is to be noted is the G.P. and the common ratio rr. A common ratio is obtained when we take the ratio of two consecutive terms of a G.P.