Trigonometric Equations

To solve the given series equation:
1+sin‌x+sin‌2x+sin‌3x+...=4+2√3
we recognize that this is an infinite geometric series with the first term a=1 and the common ratio r=sin‌x. The sum of an infinite geometric series is given by:
S=‌

a

1−r

Here, substituting the given values:
‌

1

1−sin‌x

=4+2√3
To find sin‌x, we solve:
1−sin‌x=‌

1

4+2√3

Rationalizing the denominator gives:
1−sin‌x=‌

4−2√3

4

=1−‌

√3

2

Thus:
sin‌x=‌

√3

2

Given the constraints 0<x<π and x≠‌

π

2

, we find that x could be ‌

π

3

or ‌

2π

3

. The solutions within this range are:
x=‌

π

3

,‌

2π

3