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Dima020 [189]
3 years ago
15

steven and jaden are solving b is either of them correct? explain your reasoning and in complete sentences.

Mathematics
1 answer:
sveta [45]3 years ago
4 0

Answer:

17

Step-by-step explanation:

You just add them

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Solve the equation for all real values of x.<br> cosxtanx - 2 cos x=-1
IceJOKER [234]

Solution to equationcosxtanx - 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 equationcosxtanx - 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
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