Question

(2sinX+1)(2sin²X-1)=0

Answer 1

2sinx+1=0    2sinx=-1     sinx=-1/2   x= (−1)k(−π/6)+πk,k∈Z;
2sin²x-1=0  2sin²x=1      sin²x=1/2    sinx=+-1/√2=+-√2/2                                                                                                x=(−1)k(π/4)+πk,k∈Z

Answer 2

To solve the equation (2sinX+1)(2sin²X-1)=0, we need to find the values of X that make the left-hand side of the equation equal to zero.

The equation can be written as:

(2sinX+1)(2sinX-1)(sinX+1/2) = 0

Therefore, the solutions are the values of X that make any of the factors equal to zero.

So, we have three cases to consider:

Case 1: 2sinX+1 = 0

Solving for sinX, we get:

sinX = -1/2

The solutions in the interval [0, 2π) are:

X = π/6 and X = 7π/6.

Case 2: 2sin²X - 1 = 0

Solving for sinX, we get:

sinX = ±√(1/2)

The solutions in the interval [0, 2π) are:

X = π/4, X = 3π/4, X = 5π/4, and X = 7π/4.

Case 3: sinX + 1/2 = 0

Solving for sinX, we get:

sinX = -1/2

The solutions in the interval [0, 2π) are:

X = 7π/6 and X = 11π/6.

Therefore, the solutions to the equation (2sinX+1)(2sin²X-1)=0 in the interval [0, 2π) are:

X = π/6, 7π/6, π/4, 3π/4, 5π/4, 7π/4, 11π/6.