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?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

There is no mistake in your answer but something it's missing.

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}.

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