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)

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) (Show Source): You can put this solution on YOUR website!

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

, for any integer k.
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SOLUTION: 4cosx+5cosx=4 The answer is: cosx=(4/9), right? But, how do you write this in terms of arccos &#952;? Would it be: arcoss(4/9)+2&#960;k, for any integer k, or +/-arcoss(4/9)

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)