-1

Solve $$\sqrt{3}\cos x-\sin x=\sqrt 2,\qquad 0^\circ\leq x\leq 360^\circ.$$

I try as follows.

\begin{align} &\sqrt{3}\cos x-\sin x=\sqrt 2\\ \iff & (\sqrt{3}\cos x-\sin x)^2=(\sqrt 2)^2\\ \iff & 3\cos^2 x +\sin^2 x-2\sqrt 3 \sin x\cos x=2\\ \iff & 3\left(\dfrac{1}{2}+\dfrac{1}{2}\cos2x\right) +\left(\dfrac{1}{2}-\dfrac{1}{2}\cos2x\right) x-\sqrt 3 \sin 2x=2\\ \iff & 2+\cos2x-\sqrt 3 \sin 2x=2\\ \iff & \cos2x-\sqrt 3 \sin 2x=0\\ \iff & \cos2x=\sqrt 3 \sin 2x\\ \iff & \dfrac{\sin 2x}{\cos 2x}=\dfrac{1}{\sqrt 3} \\ \iff & \tan 2x=\dfrac{\sqrt 3}{ 3} \end{align} We have \begin{align} &2x=30^\circ+k\cdot180^\circ\\ \iff&x=15^\circ+k\cdot90^\circ, k\in\mathbb Z. \end{align}

For $k=0$ we have $x=15^\circ$

For $k=1$ we have $x=105^\circ$

For $k=2$ we have $x=195^\circ$

For $k=3$ we have $x=285^\circ$.

My question:

I have four solutions for this problem. But why if I substitute $x=105^{\circ}$ and $x=195^\circ$ to the original equation the result is $\sqrt{3}\cos x-\sin x=-\sqrt2\neq \sqrt2$? What the mistake?

Ongky Denny Wijaya
  • 2,906

1 Answers1

2

You introduced spurious solutions when you used $$\sqrt{3}\cos x-\sin x=\sqrt 2 \implies (\sqrt{3}\cos x-\sin x)^2=(\sqrt 2)^2$$ and you should not have written $\iff$.

The extra solutions satisfy $\sqrt{3}\cos x-\sin x=-\sqrt 2$ as, if you square both sides of that, you again get $(\sqrt{3}\cos x-\sin x)^2=(-\sqrt 2)^2=(\sqrt 2)^2$.

Henry
  • 169,616