Question #b3c1c

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Answer 1 · 2017-04-16T18:49:41

In degrees:

$x = 150^{\circ} + n (360^{\circ})$ $x = 150^@ + n(360^@)$ and $x = 210^{\circ} + n (360^{\circ})$ $x = 210^@ + n(360^@)$ where $n in ZZ$

In radians:

$x = (5pi)/6 + n(2pi)$ and $x = (7pi)/6 + n(2pi)$ where $n in ZZ$

Given: $cos(x)=-(sqrt12)/4$

Simplify the right side:

$cos(x)=-(sqrt3)/2$

This is known to occur in both the 2nd and 3rd quadrants

In degrees:

$x = 150^@ + n(360^@)$ and $x = 210^@ + n(360^@)$ where $n in ZZ$

In radians:

$x = (5pi)/6 + n(2pi)$ and $x = (7pi)/6 + n(2pi)$ where $n in ZZ$