SOLUTION: 4cosx+5cosx=4 The answer is: cosx=(4/9), right? But, how do you write this in terms of arccos θ? Would it be: arcoss(4/9)+2πk, for any integer k, or +/-arcoss(4/9)

Algebra ->  Trigonometry-basics -> SOLUTION: 4cosx+5cosx=4 The answer is: cosx=(4/9), right? But, how do you write this in terms of arccos θ? Would it be: arcoss(4/9)+2πk, for any integer k, or +/-arcoss(4/9)      Log On


   

Question 974575: 4cosx+5cosx=4

The answer is: cosx=(4/9), right?
But, how do you write this in terms of arccos θ? Would it be: arcoss(4/9)+2πk, for any integer k, or +/-arcoss(4/9)+2π k, for any integer k, or is there just not a solution?
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
4cos%28x%29%2B5cos%28x%29=4

9cos%28x%29=4

cos%28x%29=4%2F9

x is the arc value.
4%2F9 is the cosine.
x would terminate in quadrant 1 or quadrant 4, because its cosine is positive. You would need a period of 2pi.

x=arcos%284%2F9%29%2B2k%2Api, for any integer k.