#cos(x)=-sqrt(3)/2#
If we apply the inverse of #cos#, we can isolate #x# and use a calculator to solve:
#cancel(cos^-1)(cancel(cos)(x)) = cos^-1(-sqrt(3)/2)#
#x = cos^-1(-sqrt(3)/2)#
#x=(5pi)/6#
My calculator only gave me one answer, but that may not be correct....
Now, let's try not using a calculator, and use our logic
In trigonometry, there are some ratios that keep popping up:
#sqrt(3)/2#, #1/2#, and #sqrt2/2#
Let's focus on #sqrt3/2#:
#sin(60^o)# will give you #sqrt(3)/2#, or #sin(pi/3)#.
Sine and Cosine are supplemantary, so if #sin(60)# gives you #sqrt(3)/2#, #cos(90-60)# will give you the same answer.
#Cos(30^o)# or #cos(pi/6)# equals #sqrt(3)/2#. Now we have a referance angle for our problem.
Now, we need to find the angle that gill give us the ratio #-sqrt(3)/2#. To find that, we need to find the quadrant that gives us a negative number:
#color(white)(----) color(white)(S) color(white)(--) color(black)(|) color(white)(--) color(white)(A) color(white)(----)#
#color(white)(----) color(white)(S) color(white)(--) color(black)(|) color(white)(--) color(white)(A) color(white)(----)#
#color(white)(----) color(red)(S) color(white)(--) color(black)(|) color(white)(--) color(red)(A) color(white)(----)#
#color(white)(----) color(white)(S) color(white)(--) color(black)(|) color(white)(--) color(white)(A) color(white)(----)#
#color(black)(----) color(white)(S) color(black)(--) color(black)(|) color(black)(--) color(white)(A) color(black)(----)#
#color(white)(----) color(white)(S) color(white)(--) color(black)(|) color(white)(--) color(white)(A) color(white)(----)#
#color(white)(----) color(white)(S) color(white)(--) color(black)(|) color(white)(--) color(white)(A) color(white)(----)#
#color(white)(----) color(red)(T) color(white)(--) color(black)(|) color(white)(--) color(red)(C) color(white)(----)#
#color(white)(----) color(white)(S) color(white)(--) color(black)(|) color(white)(--) color(white)(A) color(white)(----)#
#color(red)(A)# tells us that All functions are positive in this quadrant
#color(red)(S)# tells us that Sine functions are positive in this quadrant
#color(red)(T)# tells us that Tangent functions are positive in this quadrant
#color(red)(C)# tells us that Cosine functions are positive in this quadrant
I remember this using the mnemonic All Students Take Calculus
We're looking for #cos(theta) = color(red)(-)sqrt(3)/2#. That means we need to know the quadrants where the rations are negative for #cos#.
The quadrants we are looking for are Q II and III
If we go around the circle by #pi/6#:
QI
#color(blue)(pi/6)#
#(2pi)/6# or #color(blue)(pi/3)#
QII
#(3pi)/6# or #color(blue)(pi/2)#
#(4pi)/6# or #color(blue)((2pi)/30#
#color(blue)((5pi)/6)#
QIII
#(6pi)/6# or #color(blue)(pi)#
#color(blue)((7pi)/6)#
#(8pi)/6# or #color(blue)((4pi)/3)#
QIV
#(9pi)/6# or #color(blue)((3pi)/2)#
#(10pi)/6# or #color(blue)((5pi)/3)#
#color(blue)((11pi)/6)#
#(12pi)/6# or #color(blue)(2pi)#
The two solutions we care about are in Q II and III and are based on the reference angle #pi/6#
The only ones that follow those restraints are #(5pi)/6# and #(7pi)/6#
And just to check that both values are valid, let's use a calulator to make sure that #cos((5pi)/6)# and #cos((7pi)/6)# both give us #-sqrt3/2#
It does! So, both solutions are correct. Good work!
