Solving $\sqrt{3} \sin x = \cos x$ for $0

Question

So I have problem for this question, where I get a two different solutions to the same question, and is any of them wrong? Because according to my math book, I should be getting only 2 solutions, where I'm getting 4.

When $0<x<2\pi$, solve $$\sqrt{3} \sin x = \cos x$$

Here, I've squared both sides so it becomes

$$3\sin^2x = \cos^2x$$

and since $\cos^2x = 1-\sin^2x$

$$3\sin^2x = 1-\sin^2x$$

$$\sin^2x = \frac 1 4$$

which becomes

$$\sin x = \pm \frac 12$$

and $$ x = \frac \pi 6, \frac {5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}.$$

Is there anything wrong with this solution? and if what please indicate.

Answer 1

In you very first step, you took the square of both sides of your equation. This breaks the equivalence. Of course, it is true that $\sqrt{3} \sin x = \cos x$ implies $3 \sin^2 x = \cos^2x$. But, conversely, when you find a solution of the latter, it is not automatically a solution of the first.

So, starting with your list of 4 candidates, you should check which satisfy the initial equation.

Answer 2

Here is a completely different approach, which assumes that you know the angle addition formula $\sin (x-y)= \sin x \cos y - \cos x \sin y$.

\begin{align} \sqrt 3 \sin x - \cos x = 0 & \iff \frac{ \sqrt 3}{2} \sin x - \frac 12 \cos x = 0 \\ &\iff \cos \frac{\pi}{6} \sin x - \sin \frac{\pi}{6} \cos x = 0 \\& \iff \sin \left ( x - \frac {\pi}{6} \right ) = 0.\end{align}

If you also know $x \in (0, 2 \pi )$, then $x= \dfrac {\pi}{6}, \dfrac{7 \pi}{6}$ are your solutions.

As noted above, squaring both sides of an equation can introduce extraneous solutions because two unequal numbers can have equal squares.

Answer 3

$\tan$ is $\pi$-periodic and monotonically increasing, and you are looking for solutions in $[0,2\pi]$, so you should only have two solutions. \begin{align} \sqrt3\sin x&=\cos x\\ \frac{\sin x}{\cos x}&=\frac1{\sqrt3}\\ \tan x&=\frac1{\sqrt3}\\ x&=\frac\pi6,\frac{7\pi}6 \end{align}

Answer 4

The fact that $f^{2}(x)=g^{2}(x)$ does not mean that $f(x)=g(x)$. Your answer is correct up to the minus sign. To see the correct answers you can check your answers and see what are the two correct ones or you can look at the unit circle and see where does the sign of $\sin$ and $\cos$ are the same.

Answer 5

Implicit in the other comments and answers on this page is the fact that solving an equation means to find its solution set. This initially may sound tautotlogical or belabouring the obvious, but hear me out.

Now, you wouldn't agree that $\{0,4\}$ is the solution set of $$2x-6=0,\tag1$$ and would protest that it contains one extraneous solution. However, this working is perfectly mathematically valid: $$2x-6=0\\(2x-6)x=0x\\(2x-6)x=0\\x=0\;\text{or}\;3$$ sumarising the above: as $x$ varies, if $2x-6=0$ is satisfied, then $x$ can equal only $0$ or $3.$

(To be clear: the conditional's converse is not being asserted.)

At this point, we can validly assert that $\{0,4\}$ is the candidate solution set of equation $(1).$

What's going on is that when our chain of logic is tacitly purely forward—as is typically the case—obtaining an equation's actual solution set sometimes requires pruning the candidate solution set that you have created, which happens when your choice of method introduces a "non-reversible" step (e.g., the first one above).

But all is not dire: as long every step in your presentation is a true statement of the form Step X implies Step Y (this includes the first step above), requisite solutions never accidentally get discarded. So, given an equation, every candidate solution set contains every possible solution.

Please refer to this answer for a more concrete explanation.