It is 153/250. message me if u need any help. lol
Answer:
See below
Step-by-step explanation:
<u>An invalid solution.</u>
When you use the quadratic formula, for example, to solve for 't', time, you will often find one of the values of 't' to be NEGATIVE value for time.....this would be an <u>extraneous</u> solution.
Sometimes when working with LOG equations you may fin an answer which is negative.....you cannot have a LOG of a negative number...THAT solution would be EXTRANEOUS
ETC
SO, always check your answers to see if they 'work' for your problem.
The parabola hits (-2,1) and (0,-1).
As you can see from the pictures, we can immediately cross out options B and D, since both parabola open upwards.
From the pictures, we can see that the answer is C, since only that parabola opens downwards and hits the points (-2,1) and (0,-1).
U would have to find the lcm which will convert the fraction into a equalivalent one
![\bf \textit{Double Angle Identities} \\\\ sin(2\theta)=2sin(\theta)cos(\theta) \\\\ cos(2\theta)= \begin{cases} cos^2(\theta)-sin^2(\theta)\\ 1-2sin^2(\theta)\\ 2cos^2(\theta)-1 \end{cases}\quad \begin{array}{llll} \\ \leftarrow \textit{we'll use}\\ \leftarrow \textit{these two} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{double-angle identity}}{2cos^2(3x)-1}=-sin(3x)\implies cos[2(3x)]=-sin(3x)](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BDouble%20Angle%20Identities%7D%20%5C%5C%5C%5C%20sin%282%5Ctheta%29%3D2sin%28%5Ctheta%29cos%28%5Ctheta%29%20%5C%5C%5C%5C%20cos%282%5Ctheta%29%3D%20%5Cbegin%7Bcases%7D%20cos%5E2%28%5Ctheta%29-sin%5E2%28%5Ctheta%29%5C%5C%201-2sin%5E2%28%5Ctheta%29%5C%5C%202cos%5E2%28%5Ctheta%29-1%20%5Cend%7Bcases%7D%5Cquad%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5C%5C%20%5Cleftarrow%20%5Ctextit%7Bwe%27ll%20use%7D%5C%5C%20%5Cleftarrow%20%5Ctextit%7Bthese%20two%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bdouble-angle%20identity%7D%7D%7B2cos%5E2%283x%29-1%7D%3D-sin%283x%29%5Cimplies%20cos%5B2%283x%29%5D%3D-sin%283x%29)
![\bf 1-2sin^2(3x)=-sin(3x)\implies 0=\stackrel{\stackrel{ax^2+bx+c}{\downarrow }}{2sin^2(3x)-sin(3x)-1} \\\\\\ 0=[2sin(3x)+1][sin(3x)-1] \\\\[-0.35em] ~\dotfill\\\\ 0=2sin(3x)+1\implies -1=2sin(3x)\implies -\cfrac{1}{2}=sin(3x) \\\\\\ \cfrac{7\pi }{6}~,~\cfrac{11\pi }{6}=3x\implies \boxed{\cfrac{7\pi }{18}~,~\cfrac{11\pi }{18}=x} \\\\[-0.35em] ~\dotfill\\\\ 0=sin(3x)-1\implies 1=sin(3x)\implies \cfrac{\pi }{2}=3x\implies \boxed{\cfrac{\pi }{6}=x}](https://tex.z-dn.net/?f=%5Cbf%201-2sin%5E2%283x%29%3D-sin%283x%29%5Cimplies%200%3D%5Cstackrel%7B%5Cstackrel%7Bax%5E2%2Bbx%2Bc%7D%7B%5Cdownarrow%20%7D%7D%7B2sin%5E2%283x%29-sin%283x%29-1%7D%20%5C%5C%5C%5C%5C%5C%200%3D%5B2sin%283x%29%2B1%5D%5Bsin%283x%29-1%5D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%200%3D2sin%283x%29%2B1%5Cimplies%20-1%3D2sin%283x%29%5Cimplies%20-%5Ccfrac%7B1%7D%7B2%7D%3Dsin%283x%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B7%5Cpi%20%7D%7B6%7D~%2C~%5Ccfrac%7B11%5Cpi%20%7D%7B6%7D%3D3x%5Cimplies%20%5Cboxed%7B%5Ccfrac%7B7%5Cpi%20%7D%7B18%7D~%2C~%5Ccfrac%7B11%5Cpi%20%7D%7B18%7D%3Dx%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%200%3Dsin%283x%29-1%5Cimplies%201%3Dsin%283x%29%5Cimplies%20%5Ccfrac%7B%5Cpi%20%7D%7B2%7D%3D3x%5Cimplies%20%5Cboxed%7B%5Ccfrac%7B%5Cpi%20%7D%7B6%7D%3Dx%7D)
now, those angles are the angles in the range of [0, 2π] only.
a general solution angles will be
