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3 years ago

Create a set of coordinates to model a relation that is linear function.

Mathematics

1 answer:

weqwewe [10]3 years ago

3 0

Answer:

Step-by-step explanation:

The following points (coordinates) lie on a straight line with slope 3:

(1, 0), (2, 3), (3, 6), (4, 9) ... and so on

So the linear function is y = 3x + b, where b is the y-intercept.

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4 years ago

Solve the equation for all real values of x.<br> cosxtanx - 2 cos x=-1

IceJOKER [234]

Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

<u>Step-by-step explanation:</u>

Here we have , $cosxtanx - 2 cos^2 x=-1$ . Let's solve :

⇒ $cosxtanx - 2 cos^2 x=-1$

⇒ $cosx(\frac{sinx}{cosx}) - 2 cos^2 x=-1$

⇒ $sinx = 2 cos^2 x-1$

⇒ $sinx = 2 (1-sin^2x)-1$

⇒ $sinx = 1-2sin^2x$

⇒ $2sin^2x+sinx-1=0$

By quadratic formula :

⇒ $sinx = \frac{-b \pm \sqrt{b^2-4ac} }{2a}$

⇒ $sinx = \frac{-1 \pm \sqrt{1^2-4(2)(-1)} }{2(2)}$

⇒ $sinx = \frac{-1 \pm3}{4}$

⇒ $sinx = \frac{1}{2} , sinx =-1$

⇒ $sinx = sin\frac{\pi}{6} , sinx = sin\frac{3\pi}{2}$

⇒ $x=\frac{\pi}{6} , x=\frac{3\pi}{2}$

But at $x=\frac{3\pi}{2}$ we have equation undefined as $cos\frac{3\pi}{2}=0$ . Hence only solution is :

⇒ $x=\frac{\pi}{6}$

Since , $sin(\pi -x)=sinx$

⇒ $x=\pi -\frac{\pi}{6} = \frac{5\pi}{6}$

Now , General Solution is given by :

⇒ $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$

Therefore , Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

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