Answer:
[tex]x=(-1)^{k+1}\cdot \dfrac{\pi}{6}+\pi k,\ k\in Z[/tex]
Step-by-step explanation:
Consider the equation
[tex]2\sin^2x-5\sin x-3=0[/tex]
Substitute
[tex]u=\sin x[/tex]
into the equation:
[tex]2u^2-5u-3=0[/tex]
This is quadratic equation, where
[tex]D=(-5)^2-4\cdot 2\cdot (-3)=25+24=49\\ \\\sqrt{D}=7[/tex]
Hence,
[tex]u_1=\dfrac{-(-5)-\sqrt{D}}{2\cdot 2}=\dfrac{5-7}{4}=-\dfrac{1}{2}\\ \\u_2=\dfrac{-(-5)+\sqrt{D}}{2\cdot 2}=\dfrac{5+7}{4}=3[/tex]
Now, use the substitution again:
1. When [tex]u_1=-\dfrac{1}{2},[/tex] then
[tex]\sin x=-\dfrac{1}{2}\\ \\x=(-1)^k\sin^{-1}\left(-\dfrac{1}{2}\right)+\pi k,\ k\in Z\\ \\x=(-1)^k\cdot \left(-\dfrac{\pi}{6}\right)+\pi k,\ k\in Z\\ \\x=(-1)^{k+1}\cdot \dfrac{\pi}{6}+\pi k,\ k\in Z[/tex]
2. When [tex]u_2=3,[/tex] then
[tex]\sin x=3[/tex]
This equation has no solutions, because the range of [tex]\sin x[/tex] is [tex][-1,1][/tex] and 3>1.
