JEE Advanced 2020 Concept Recapitulation Test 1 Paper 1

The conditions for the existence of a solution are
1. 2‌cos‌2‌x−1≠0⇒cos‌2‌x≠‌

1

2

⇒2x≠‌

π

6

2. tan‌x≠0⇒x≠nπ
3. cos2x−3sin2x≠0⇒tan2x≠‌

1

3

⇒x≠±‌

π

6

As 2‌cos‌x−1=2(cos2x−sin2x)−1
=2(cos2x−sin2x)−(cos2x+sin2x)
=cos2x−3sin2x
So the given equation can be written as b‌sin‌x=b+sin‌x
⇒‌‌sin‌x=‌

b

b−1

Which is possible if −1≤‌

b

b−1

≤1
⇒‌‌‌

2b−1

b−1

≥0 and ‌

1

b−1

≤0
⇒‌‌2b−1≤0⇒b≤‌

1

2

But for b=‌

1

2

,sin‌x=−1
which is not possible as x≠−‌

π

2

.
Hence equation has a solution if b<‌

1

2

.