At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
[tex]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>[/tex]
[tex]sinx=1\ and\ x\in\left< 0;\ 2\pi\right)\\\\then\ x=\frac{\pi}{2}.[/tex]
[tex]sinx=1\ and\ x\in\left< 0;\ 2\pi\right)\\\\then\ x=\frac{\pi}{2}.[/tex]
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.