Question

The number of solutions of the equation 2sin?x + sinx - 1 = 0) in (0,2π) is.
1
1
2
2
قی)
3
4
4​

Answer 1

Answer:

Given, 1+sinx⋅sin

2

2

x

=0

⇒2+2sinx⋅sin

2

2

x

=0

⇒2+sinx(1−cosx)=0

⇒4+2sinx(1−cosx)=0

⇒4+2sinx−sin2x=0

⇒sin2x=2sinx+4

Above is not possible for any value of x as LHS has maximum value 1 and RHS has minimum value 2.

Hence, there is no solution.