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=(-b-\sqrt\Delta)/(2a)\to t_1=(-3-√(49))/(2\cdot2)=(-3-7)/(4)=(-10)/(4)\\otin\left< -1;\ 1\right>\\\\t_2=(-b+\sqrt\Delta)/(2a)\to t_2=(-3+√(49))/(2\cdot2)=(-3+7)/(4)=(4)/(4)=1\in\left < -1;\ 1 \right>`


`sinx=1\ and\ x\in\left< 0;\ 2\pi\right)\\\\then\ x=(\pi)/(2).`