Question

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

2sin2x+sinx=1

Answer 1

`(\pi)/(6), (5\pi)/(6), (3\pi)/(2)`

Explanation:

Given the quadratic equation as:

`2sin^2x+sinx=1\\OR\\2sin^2x+sinx-1=0`

Let us put
`sinx=y` for simplicity of the equation:

Now, the equation becomes:

`2y^2+y-1=0`

Now, let us try to solve the quadratic equation:

`\Rightarrow 2y^2+2y-y-1=0\\\Rightarrow 2y(y+1)-1(y+1)=0\\\Rightarrow (2y-1)(y+1)=0\\\Rightarrow 2y-1 = 0, y+1 = 0\\\Rightarrow y = (1)/(2), y = -1`

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

`sinx = (1)/(2)`

and

`sinx=-1`

Let us try to find the values of
`x` in the interval
`[0, 2\pi)`.

`sin\theta` can have a value equal to
`(1)/(2)` in 1st and 2nd quadrant.

So,
`x` can be

`30^\circ, 150^\circ\\OR\\(\pi)/(6), (5\pi)/(6)`

For
`sinx=-1`,

`x = 270^\circ\ or\ (3 \pi)/(2)`

So, the answer is:

`(\pi)/(6), (5\pi)/(6), (3\pi)/(2)`