How do you solve $ln (cos x) = 0$ $ln(cosx)=0$ in the interval [0,360]?

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Answer 1 · 2016-08-20T12:46:01

$x = 0$ $x= 0$ or $360$ $360$ for $x \in (0, 360)$ $x in (0, 360)$

$ln (cos x) = 0$ $ln(cosx) =0$

$cos x = e^{0} \to cos x = 1$ $cosx = e^0 -> cosx=1$

Hence: $x = arccos (1)$ $x=arccos(1)$

Since: $arccos (1) = 0$ $arccos(1) = 0$ and $2 π \in (0, 2 π)$ $2pi in (0, 2pi)$

Therefore, in degrees: $x = 0$ $x= 0$ or $360$ $360$ for $x \in (0, 360)$ $x in (0, 360)$

How do you solve ln(cosx)=0ln(cosx)=0ln(cosx)=0 in the interval [0,360]?