Question #90e09

1 Answer
Noah G
Nov 27, 2017

I would let #u = cosx#.

#10u^2 +13u + 4 = 0#

We can readily solve this by factoring.

#10u^2 + 5u + 8u + 4 = 0#

#5u(2u + 1) + 4(2u + 1) = 0#

#(5u + 4)(2u + 1) = 0#

#u = -4/5 or -1/2#

Reverse your substitution.

#cosx =-4/5 or cosx = -1/2#

I'll leave the rest up to you. What you need to understand is that cosine is negative in the 2nd and 3rd quadrants. Find the value of #arccos(4/5)# using a calculator and then add that result to #180# and subtract it from #180#. Repeat for the second solution.

Hopefully this helps!

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