# How do you solve tanxsinx-tanx=0 in the interval [0,360]?

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Frenst Share
Feb 20, 2017

$\tan \left(x\right) \cdot \sin \left(x\right) - \tan \left(x\right) = 0$ in the interval [0;360] for x={0;180;360}

#### Explanation:

• Factorize: $\tan \left(x\right) \cdot \sin \left(x\right) - \tan \left(x\right) = 0$

$\tan \left(x\right) \cdot \left(\sin \left(x\right) - 1\right) = 0$

• Determinate the domain of the function :

the domain of function $\tan \left(x\right)$ is $\mathbb{R} - \left\{\frac{\Pi}{2} + k \Pi\right\}$ with $k \in \mathbb{Z}$
so the domain of function $\tan \left(x\right)$ in the interval [0;360] is [0;90[nn]90;270[nn]270;360]

the domain of function $\sin \left(x\right)$ is $\mathbb{R}$
so the domain of function $\tan \left(x\right)$ in the interval [0;360] is [0;360]

so the domain of function $\tan \left(x\right) \cdot \left(\sin \left(x\right) - 1\right) = 0$ in the interval [0;360] is the union of the domain of function $\tan \left(x\right)$ and function $\sin \left(x\right) - 1$
so it's [0;90[nn]90;270[nn]270;360]
or in an other way to write it [0;360]-{90;270}

• find the zero of the function in this domain :

$\tan \left(x\right) \cdot \left(\sin \left(x\right) - 1\right) = 0$ in the interval [0;360]-{90;270} when :
- $\tan \left(x\right) = 0$ in the interval [0;360]-{90;270}
- or $\sin \left(x\right) - 1 = 0$ in the interval [0;360]-{90;270}

$\tan \left(x\right) = 0$ in the interval [0;360]-{90;270} for x={0;180;360}

$\sin \left(x\right) - 1 = 0$
$\sin \left(x\right) = 1$ in the interval [0;360] for x={90;270}
$\sin \left(x\right) = 1$ in the interval [0;360]-{90;270} for $x = \emptyset$

so $\tan \left(x\right) \cdot \sin \left(x\right) - \tan \left(x\right) = 0$ in the interval [0;360] for x={0;180;360}

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer
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