Finding general solution for trigonometric equation

[Vertciel]

15. Solve the equation $\displaystyle \sin^2 x - \sqrt{2} \cos x = \cos^2 x + \sqrt{2}\cos x + 2$ for x in the interval $\displaystyle 0 \leq x \leq 2\pi$. Then, write a general solution for this equation.

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Hello there,

My general solution differs from the one given by the textbook by not having a negative sign. Could anyone please explain why the textbook's general solution includes only solutions which lie to the right of $\displaystyle x=\frac{3\pi}{4}$?

Thank you.

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$\displaystyle \sin^2 x - \sqrt{2} \cos x = \cos^2 x + \sqrt{2}\cos x + 2$

$\displaystyle -2\cos^2 x - 2\sqrt{2}\cos x - 1 = 0$

$\displaystyle \Rightarrow \cos x = \frac{-\sqrt{2}}{2} \Rightarrow x = \frac{3\pi}{4}$

Therefore,

$\displaystyle x = 2n\pi \pm \frac{3\pi}{4}$

However, textbook gives:

$\displaystyle x = 2n\pi + \frac{3\pi}{4}$

 

[Loren]

Did you check it? It looks to me like the negative root is extraneous.

 

[Vertciel]

Thanks for your reply, Loren.

My teacher says that for the general solution of trigonometric equations with the $\displaystyle \cos$ function,

$\displaystyle \theta = 2n\pi \pm \cos^{-1} (k)$, if $\displaystyle \cos x = k$.

Therefore, is my teacher's definition redundant?

 

[Deleted member 4993]

15. Solve the equation $\displaystyle \sin^2 x - \sqrt{2} \cos x = \cos^2 x + \sqrt{2}\cos x + 2$ for x in the interval $\displaystyle 0 \leq x \leq 2\pi$. Then, write a general solution for this equation.

----

Hello there,

My general solution differs from the one given by the textbook by not having a negative sign. Could anyone please explain why the textbook's general solution includes only solutions which lie to the right of $\displaystyle x=\frac{3\pi}{4}$?

Thank you.

----
$\displaystyle \sin^2 x - \sqrt{2} \cos x = \cos^2 x + \sqrt{2}\cos x + 2$

$\displaystyle -2\cos^2 x - 2\sqrt{2}\cos x - 1 = 0$

$\displaystyle \Rightarrow \cos x = \frac{-\sqrt{2}}{2} \Rightarrow x = \frac{3\pi}{4}$

Therefore,

$\displaystyle x = 2n\pi \pm \frac{3\pi}{4}$

However, textbook gives:

$\displaystyle x = 2n\pi + \frac{3\pi}{4}$

I believe books answer is incorrect. The domain is given to be 0 ? 2?. So the answer should not have 2n?.

Your answer is correct 2n? ± 3?/4. You have two solutions in the given domain ? 3?/4 & 5?/4 (which is 2? - 3?/4)