Solve equation 3sec^2 x = 5 + tanx (confusing answer)

[Monkeyseat]

Question:

Solve the equation 3sec^2 x = 5 + tanx, giving all solutions in the interval -180 < x < 180 degrees.

Working:

3(1 + tan^2 x) = 5 + tanx
3 + 3tan^2 x = 5 + tanx
3tan^2 x - tanx - 2 = 0
(3tanx + 2)(tanx - 1) = 0

So:

3tanx + 2 = 0
tanx = -2/3
From drawing a CAST diagram, x = 146.3, -33.7 degrees.

and

tanx - 1 = 0
tanx = 1
From drawing a CAST diagram, x = 45, -135 degrees.

Therefore, all values for x are -135, -33.7, 45, 146.3 degrees.

I thought that was correct, but yet again the book says otherwise. The book says: 45, 146.3, 225, 326.3 degrees. Unless I've missed something, is that even possible if the solutions have to be in the interval -180 < x < 180 degrees? It definitely says this I have checked.

Have I done something wrong or has the book just done it for the wrong interval (by the looks of it 0 < x <360 degrees)?

Thanks again.

 

[mmm4444bot]

Therefore, rounded to one decimal point, all values for x are -135º, -33.7º, 45º, and 146.3º.

Correct!

the book says [all positive angles]... is that even possible if the solutions have to be in the interval -180 < x < 180 degrees? Clearly not!

 

[Deleted member 4993]

If you have any choice in the subject - change that confounding book. Too many mistakes....

 

[Monkeyseat]

Thanks for clarifying. Unfortunately, I can't change the book Subhotosh Khan - it's the one everyone has to use...

 

[PAULK]

Question:

Solve the equation 3sec^2 x = 5 + tanx, giving all solutions in the interval -180 < x < 180 degrees.

Working:

3(1 + tan^2 x) = 5 + tanx
3 + 3tan^2 x = 5 + tanx
3tan^2 x - tanx - 2 = 0
(3tanx + 2)(tanx - 1) = 0

So:

3tanx + 2 = 0
tanx = -2/3
From drawing a CAST diagram, x = 146.3, -33.7 degrees.

and

tanx - 1 = 0
tanx = 1
From drawing a CAST diagram, x = 45, -135 degrees.

Therefore, all values for x are -135, -33.7, 45, 146.3 degrees.

I thought that was correct, but yet again the book says otherwise. The book says: 45, 146.3, 225, 326.3 degrees. Unless I've missed something, is that even possible if the solutions have to be in the interval -180 < x < 180 degrees? It definitely says this I have checked.

Have I done something wrong or has the book just done it for the wrong interval (by the looks of it 0 < x <360 degrees)?

Thanks again.

You have just discovered that the brilliant but lazy professor who wrote the text and the talented but overworked and underpaid grad student who wrote the answers are not one and the same.