# I'm trying to find all solutions to the equation 5cos(x+3)=1 from 0 to 2pi and can't figure out why my answer isn't matching the back of the book.I've divided both sides by 5 getting cos(x+3) = 1/5...

I'm trying to find all solutions to the equation 5cos(x+3)=1 from 0 to 2pi and can't figure out why my answer isn't matching the back of the book.

I've divided both sides by 5 getting cos(x+3) = 1/5 then did inverse cos of 1/5 leaving x+3 = invcos 1/5 then subtracted 3 from that leaving me an angle. Where am i going wrong?

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To solve an elementary trigonomteric equation involving cosine function, you'll have to do the following steps:

cos x = a

x = `+-` arccos (a) + 2k*`pi` , where k`in` Z, the set of integer numbers

Therefore, we'll solve the equation from the point:

cos(x+3) = 1/5

x + 3 = `+-` arccos(1/5) + 2k*`pi`

x = `+-` arccos(1/5)  + 2k`pi`

We'll consider k =0 => 2k`pi` = 0

arccos (1/5) = arccos(0.2) = 0.064 `pi`

x = 0.064 `pi` - 3

If we'll consider pi = 3.14, we'll get:

x = 0.20096 - 3

x = -2.79 approx.

x = - 0.064 `pi` - 3

x = -3.20 approx.

We'll consider k =1 => 2k`pi` = 2`pi`

x = 0.064`pi` + 2`pi` - 3

x = 3.48 approx.

x = -0.064 `pi` + 2`pi` - 3

x = 3.07 approx

Since the angle is included in the interval [0,2`pi` ], then k cannot be larger than 1.

Therefore, the solutions of the equation, in the interval [0,2`pi ` ], for `pi` = 3.14, are: {-3.20 ; -2.79 ; 3.07 ; 3.48}.