C2 trig

You don't need cos(-x)=cos(x) here, do you know how to solve things like cos(t)=0.5? If so then just do that with t=x+40.

x+40=arcos(-0.5)

Not sure what arcos is

arcos is inverse of cos, so cos^(-1) on your calculator.

It helps to either use the quadrant method or - the one I use - the graph method.

First, graph cos(x+40), or cosx might be easier. Then solve as simultaneous equations with y = cosx and y = -0.5. The points are the intersections/solutions. You should be able to use the symmetry of cosine to calculate the other answers within the range.

When you input inverse cosine of -0.5, you'll get one solution. This is 120 degrees, which is helpful, considering is it 30 degrees to the right of the first x intercept.

Solve the equation $cos(x+40°)=-0.5$ for -180<x<180

I remember the teacher saying something about cos(-x)=cos(x) but when I put the 2 in the calculator I get different answers.


What value do I use?

Ignore the negative sign at first and find arcos(0.5)=60
Then the negative sign tells us that our angle must be in the second or third quadrant so it is 120 or 240, but only 120 will give a solution in the required range.Does the attached file help?