The following equation involves a trigonometric equation in quadratic form. Solve the equation on the interval [0,2π). 2sin2x+sinx=1

Step-by-step explanation:

Given the quadratic equation as:

Let us put for simplicity of the equation:

Now, the equation becomes:

Now, let us try to solve the quadratic equation:

So, the solution to the given trigonometric quadratic equation is:

and

Let us try to find the values of in the interval .

can have a value equal to in 1st and 2nd quadrant.

So, can be

For ,

So, the answer is:

the answer is two. hope this !