Write two numbers that multiply to the value on top and add to the value on bottom.

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

3 0

3 years ago

Why is it helpful to write numbers in different ways

Goshia [24]

You should write numbers in as many ways as you possibly can to make new connections in your brain. Knowing how to write numbers in many different ways can help you solve complex problems more easily. Doing this can also reinforce the mathematical principles and logic you have memorised.

Writing one in many different ways:

1=1/1=2/2=3/3=4/4=(-1)/(-1)=(-2)/(-2)

=1.0=1.00=1.000=(1/2)+(1/2)=(1/3)+(1/3)+(1/3)

=(1/4)+(1/4)+(1/4)+(1/4)

Writing a half in many different ways:

1/2=(1/4)+(1/4)=(1/6)+(1/6)+(1/6)

=(1/8)+(1/8)+(1/8)+(1/8)=4*(1/8)

=2/4=3/6=4/8=5/10=0.5=0.50

etc...etc...

6 0

3 years ago

Objects A and B are in harmonic motion modeled by

Zepler [3.9K]

Answer:

The phase difference between yA and yB is $\frac{-\pi }{12}$