Solution to equation
for all real values of x is
.
<u>Step-by-step explanation:</u>
Here we have ,
. Let's solve :
⇒ ![cosxtanx - 2 cos^2 x=-1](https://tex.z-dn.net/?f=cosxtanx%20-%202%20cos%5E2%20x%3D-1)
⇒ ![cosx(\frac{sinx}{cosx}) - 2 cos^2 x=-1](https://tex.z-dn.net/?f=cosx%28%5Cfrac%7Bsinx%7D%7Bcosx%7D%29%20-%202%20cos%5E2%20x%3D-1)
⇒ ![sinx = 2 cos^2 x-1](https://tex.z-dn.net/?f=sinx%20%3D%202%20cos%5E2%20x-1)
⇒ ![sinx = 2 (1-sin^2x)-1](https://tex.z-dn.net/?f=sinx%20%3D%202%20%281-sin%5E2x%29-1)
⇒ ![sinx = 1-2sin^2x](https://tex.z-dn.net/?f=sinx%20%3D%201-2sin%5E2x)
⇒ ![2sin^2x+sinx-1=0](https://tex.z-dn.net/?f=2sin%5E2x%2Bsinx-1%3D0)
By quadratic formula :
⇒ ![sinx = \frac{-b \pm \sqrt{b^2-4ac} }{2a}](https://tex.z-dn.net/?f=sinx%20%3D%20%5Cfrac%7B-b%20%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%20%7D%7B2a%7D)
⇒ ![sinx = \frac{-1 \pm \sqrt{1^2-4(2)(-1)} }{2(2)}](https://tex.z-dn.net/?f=sinx%20%3D%20%5Cfrac%7B-1%20%5Cpm%20%5Csqrt%7B1%5E2-4%282%29%28-1%29%7D%20%7D%7B2%282%29%7D)
⇒ ![sinx = \frac{-1 \pm3}{4}](https://tex.z-dn.net/?f=sinx%20%3D%20%5Cfrac%7B-1%20%5Cpm3%7D%7B4%7D)
⇒ ![sinx = \frac{1}{2} , sinx =-1](https://tex.z-dn.net/?f=sinx%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2C%20sinx%20%3D-1)
⇒ ![sinx = sin\frac{\pi}{6} , sinx = sin\frac{3\pi}{2}](https://tex.z-dn.net/?f=sinx%20%3D%20sin%5Cfrac%7B%5Cpi%7D%7B6%7D%20%2C%20sinx%20%3D%20sin%5Cfrac%7B3%5Cpi%7D%7B2%7D)
⇒ ![x=\frac{\pi}{6} , x=\frac{3\pi}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5Cpi%7D%7B6%7D%20%2C%20x%3D%5Cfrac%7B3%5Cpi%7D%7B2%7D)
But at
we have equation undefined as
. Hence only solution is :
⇒ ![x=\frac{\pi}{6}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5Cpi%7D%7B6%7D)
Since , ![sin(\pi -x)=sinx](https://tex.z-dn.net/?f=sin%28%5Cpi%20-x%29%3Dsinx)
⇒ ![x=\pi -\frac{\pi}{6} = \frac{5\pi}{6}](https://tex.z-dn.net/?f=x%3D%5Cpi%20-%5Cfrac%7B%5Cpi%7D%7B6%7D%20%3D%20%5Cfrac%7B5%5Cpi%7D%7B6%7D)
Now , General Solution is given by :
⇒ ![x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}](https://tex.z-dn.net/?f=x%3D2k%5Cpi%20%2B%20%5Cfrac%7B%5Cpi%7D%7B6%7D%20%20%2C%20x%3D2k%5Cpi%20%2B%20%5Cfrac%7B5%5Cpi%7D%7B6%7D)
Therefore , Solution to equation
for all real values of x is
.
The best answer to go with is b you’re welcome have a great day
Assuming no border or waste, 9m by 16 m translates to 29.52' x 52.49'.
use 32/8=4 sheets of plywood to cover the width on the 8' side.
For the length, use 52/4=13 pieces on the 4' side.
This gives a total of 4*13=52 sheets.
The extra 0.49' x 29.52 can be covered using the extra from the last four sheets for the width.
Check:
Area to be covered: 29.52*52.49=1550 sq. ft.
Area of plywood supplied = 52*4'x8'= 1664 sq. ft, > 1550 sq.ft.