Question

Solve the equation on the interval [0,2pi). 2sin^2x=-3sinx+5

Answer 1

$2sin^2x=-3sinx+5\\\\2sin^2x+3sinx-5=0\\\\let\ t=sinx\ then\ t\in\left< -1;\ 1 \right>\\\\2t^2+3t-5=0\\\\a=2;\ b=3;\ c=-5\\\\\Delta=b^2-4ac\to\Delta=3^2-4\cdot2\cdot(-5)=9+40=49\\\\t_1=\frac{-b-\sqrt\Delta}{2a}\to t_1=\frac{-3-\sqrt{49}}{2\cdot2}=\frac{-3-7}{4}=\frac{-10}{4}\notin\left< -1;\ 1\right>\\\\t_2=\frac{-b+\sqrt\Delta}{2a}\to t_2=\frac{-3+\sqrt{49}}{2\cdot2}=\frac{-3+7}{4}=\frac{4}{4}=1\in\left < -1;\ 1 \right>$

$sinx=1\ and\ x\in\left< 0;\ 2\pi\right)\\\\then\ x=\frac{\pi}{2}.$