Question

Solve: tan(x)(tan x+1)=0.

Please explain. Thank you!

Answer 1

Do you need to solve algebraically or graphically?

Answer 2

@swagmaster47 I am not sure what I have attached is all they gave me to work with?

Answer 3

I guess algebraically

Answer 4

The problem is that I don't see the right answer =|
I'll explain what I mean.
$$
tan(x) \cdot (tan(x) + 1) = 0
$$Means that either

\(tan(x) = 0\) or \(tan(x) + 1 = 0\)
Now since:
$$
tan(x) = \frac{sin(x)}{cos(x)}
$$We can say:
$$
tan(x) = 0 \implies \frac{sin(x)}{cos(x)} = 0 \implies sin(x) = 0 \\
x = \pi n
$$
For the other part we say:
$$
tan(x) +1 = 0 \implies tan(x) = -1 \\
\frac{sin(x)}{cos(x)} = -1 \implies sin(x) = -cos(x)
$$So basically we need the angles where \(sin(x)\) and \(cos(x)\) values have the same magnitude but opposite signs.
At \(45^\circ\) which is \(\frac{\pi}{4}\) we have \(sin(x) = cos(x) = \frac{\sqrt{2}}{2}\)
Every \(\frac{\pi}{2}\) we add or subtract from an angle will cause one of the values to swap signs.
So by subtracting \(\frac{\pi}{2}\) we can get to
$$
\frac{\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4} - \frac{2\pi}{4} = -\frac{\pi}{4}
$$And now at this angle we can see that \(sin(x)\) is negative while \(cos(x)\) remains positive, and the ratio ends up to -1 as required.
If adding/subtracting \(\frac{\pi}{2}\) once swapped the sign, then doing so twice will end up keeping the sign. So we can get the other possible solutions by:
$$
x = -\frac{\pi}{4} \pm \pi n
$$The closest solution you have is D, but it is wrong because it's missing the minus at the beginning.
The others cannot be correct too because what is added to the angle is in multiples of \(2\pi\), but it misses half of the solutions as multiples of \(\pi\) will also give correct solutions.
As a resource you can see at wolfram alpha the same kind of solution:
http://www.wolframalpha.com/input/?i=tan%28x%29%28tan%28x%29+%2B+1%29+%3D+0