Question #b3c1c

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Answer 1 · 2017-04-16T18:49:41

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#

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#