Answer:
[tex]\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}[/tex]
Step-by-step explanation:
Given the quadratic equation as:
[tex]2sin^2x+sinx=1\\OR\\2sin^2x+sinx-1=0[/tex]
Let us put [tex]sinx=y[/tex] for simplicity of the equation:
Now, the equation becomes:
[tex]2y^2+y-1=0[/tex]
Now, let us try to solve the quadratic equation:
[tex]\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 = \dfrac{1}{2}, y = -1[/tex]
So, the solution to the given trigonometric quadratic equation is:
[tex]sinx = \dfrac{1}{2}[/tex]
and
[tex]sinx=-1[/tex]
Let us try to find the values of [tex]x[/tex] in the interval [tex][0, 2\pi)[/tex].
[tex]sin\theta[/tex] can have a value equal to [tex]\frac{1}{2}[/tex] in 1st and 2nd quadrant.
So, [tex]x[/tex] can be
[tex]30^\circ, 150^\circ\\OR\\\dfrac{\pi}{6}, \dfrac{5\pi}{6}[/tex]
For [tex]sinx=-1[/tex],
[tex]x = 270^\circ\ or\ \dfrac{3 \pi}{2}[/tex]
So, the answer is:
[tex]\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}[/tex]
