Hey, I'm having trouble with these general solution questions and I'm not sure why.
sinx+cosx=1
I squared both sides and ended up with sin2x=-1, then I wasn't really sure about my next stops because the reference angle is either not acute or negative. But I guessed and did x=-pi/4. So then I got x=-npi + (-1)^n*pi/4. However there are actually 2 answers. Help please?
(sqr3 cosx)-sinx=1
I did the auxiliary angle method and got sin(x-pi/3) = 1/2, so x-pi/3 = npi + (-1)^n*pi/6, and then I added pi/3 to both sides to get x=4npi/3 + (-1)^n*pi/6. However the answer is 2pin +- pi/6. Not sure how to get that.
To resolve the problem with the first question firstly, note that the general solution to \( \sin x = \alpha\) is \( x=n\pi + (-1)^n\sin^{-1}\alpha\).
\(\sin^{-1}\alpha\) takes values between \( -\frac\pi2 \leq \sin^{-1}\alpha \leq \frac\pi2\). This is due to the range of the function \(f(x) = \sin^{-1}x\). So therefore when dealing with the general solution for \(\sin\) with negative RHS's,
negative angles are valid. Therefore solving \( \sin 2x = -1 \) would give \( \boxed{2x = n\pi + (-1)^n \left( -\frac{\pi}{2} \right)} \).
Which then becomes \(x = \frac{n\pi}{2} + (-1)^{n+1} \frac\pi4 \). Which is (possibly surprisingly)
wrong.
The dangerous thing you did was "square both sides". It may seem surprising, but squaring (especially in the context of trigonometric equations) typically introduces
new solutions to the equations that we didn't have before. Your general solution now has the values \(x = 0, \pm \frac\pi2, \pm \frac{2\pi}{2}, \pm \frac{3\pi}{2}, \pm \frac{4\pi}{2}, \pm \frac{5\pi}{2}, \pm \frac{6\pi}{2}\) and so on, when the original question
only has the solutions \( x=0, \pm \frac\pi2, \frac{4\pi}{2}, \frac{5\pi}{2},\) etc.
It is recommended to use the auxiliary angle method for the first one as well, and deal with \( \sqrt{2} \sin \left( x + \frac\pi4 \right) = 1 \) instead.
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For your second one, you should've obtained \( \sin x - \sqrt{3}\cos x = -1 \). This would've given you \(2\sin \left( x - \frac\pi3 \right) = -1 \), and hence \(x - \frac\pi3 = n\pi + (-1)^n \left( -\frac\pi6\right) \).
Which of course, does not look like what the answers had. The thing is, your approach (once you fix the sign error) is equally valid. Basically, they chose to use the other auxiliary angle formula instead:
\[ a\cos x + b\sin x = R\cos (x-\alpha)\\ \text{where }R=\sqrt{a^2+b^2}\text{ and }\tan \alpha = \frac{b}{a}. \]
If you try doing it that way instead, you'll find your answer matches their's. In the exam, either they will specify which to use, or both will be marked correct.