BITSAT 2024 Shift 1 Paper

Given the equation:
sin‌9θ=sin‌θ
This can be rewritten as:
sin‌9θ−sin‌θ=0
Using the identity sin‌C−sin‌D=2‌cos(‌

C+D

2

)‌s‌i‌n‌(‌

C−D

2

), we have:
2‌cos(‌

9θ+θ

2

)‌s‌i‌n‌(‌

9θ−θ

2

)=0
Simplifying, this becomes:
2‌cos‌5‌θ‌s‌i‌n‌4θ=0
This implies either:
‌cos‌5‌θ=0
‌sin‌4θ=0
For cos‌5‌θ=0, we have:
5θ=(2n+1)‌

π

2

This gives:
θ=(2n+1)‌

π

10

For sin‌4θ=0, we have:
4θ=nπ
This gives:
θ=‌

nπ

4

Calculating these solutions within the interval [0,2π] :
From θ=(2n+1)‌

π

10

:
θ=‌

π

10

,‌

3π

10

,‌

π

2

,‌

7π

10

,‌

9π

10

,‌

11π

10

,‌

13π

10

,‌

3π

2

,‌

17π

10

,‌

19π

10

From θ=‌

nπ

4

:
θ=0,‌

π

4

,‌

π

2

,‌

3π

4

,π,‌

5π

4

,‌

3π

2

,‌

7π

4

,2π
The total number of solutions counting common terms ‌

π

2

and ‌

3π

2

only once, is:
17