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I am required to solve the following equation:

$$\sin x \cos x = \frac{1}{2}$$

My attempt:

Rewriting $\cos x$ $$\sin x \sqrt{1 - \sin^2 x} = \frac{1}{2}$$ Squaring both sides $$\bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2$$ $$\sin^2 x (1 - \sin^2 x) = \frac{1}{4}$$

Expanding left side and multiplying both sides by 4 $$\sin^2 x - \sin^4 x = \frac{1}{4}$$ $$4\sin^2 x - 4\sin^4 x = 1$$ $$4\sin^2 x - 4\sin^4 x -1 = 0$$

Reordering left side

$$- 4\sin^4 x + 4\sin^2 x -1 = 0$$ $$4\sin^4 x - 4\sin^2 x + 1 = 0$$

Expression above can be factored as $$(2\sin^2 x - 1)(2\sin^2 x - 1) = 0$$ $$(2\sin^2 x - 1)^2 = 0$$

It follows that

$$2\sin^2 x - 1 = 0 $$ $$\sin^2 x = \frac{1}{2} $$ $$\sin x = ± \frac{1}{\sqrt{2}} $$

So the resulting angles are: $45^{\circ},135^{\circ},225^{\circ},315^{\circ}$

Is my solution correct?


The reason why I am asking is, the author of the book used different method, and the end result he got was: $$\sin2x = 1$$ So $2x = \sin^{-1}(1) = 90^{\circ},450$, and thus $x = 45^{\circ},225^{\circ}$

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    $\begingroup$ Your solution is not correct. $\sin(135^o)\cos(135^o)=\color{red}-\dfrac12$ $\endgroup$ Commented Aug 5, 2019 at 16:55
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    $\begingroup$ Since you squared a radical, you must check for extraneous solutions which are 135° and 315° in this case. At x = 135° and 315°, sinx • cosx = -1/2 $\endgroup$ Commented Aug 5, 2019 at 16:57

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Your solution is incorrect, as others pointed out. The reason is because you've made two "irreversible" operations, and did not took care of the branches they create.

The first one is:

$$ \sin x \cos x = \frac{1}{2} \Rightarrow \sin x \sqrt{1 - \sin^2 x} = \frac{1}{2} $$ This is is simply wrong, because if your are working on real numbers, the outcome of the square root must be a non-negative number. The correct statement would actually be: $$ \sin x |\cos x| = \frac{1}{2} \Rightarrow \sin x \sqrt{1 - \sin^2 x} = \frac{1}{2} $$

The second one is: $$ \sin x \sqrt{1 - \sin^2 x} = \frac{1}{2} \Rightarrow \bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2 $$

This is actually right, but you needed the arrow to go both ways, and it doesn't because: $$ \sin x \sqrt{1 - \sin^2 x} = -\frac{1}{2} \Rightarrow \bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(-\frac{1}{2}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2 $$

Because of this step, you've introduced 2 additional solutions.

You could have checked all the answers you've obtained to verify which remained valid, but the advantage of the double-arc method is that you circumvent those irreversible operations, making it easier to reach the final answer.

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Use that $$\sin(2x)=2\sin(x)\cos(x)$$

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    $\begingroup$ The question is tagged with [proof-verification], and OP specifically asks if their solution is correct, and why it is different from the solution in the book. Your answer just repeats the solution from the book. $\endgroup$
    – Martin R
    Commented Aug 5, 2019 at 17:26
  • $\begingroup$ The others also, dear Martin R $\endgroup$ Commented Aug 5, 2019 at 17:27
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    $\begingroup$ That is a poor excuse from a >80 k user. – Btw, I see several answers explaining why OP's solution is not correct. $\endgroup$
    – Martin R
    Commented Aug 5, 2019 at 17:32
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$$2\sin(x)\cos(x)=1$$ $$\sin(2x)=1$$ $$2x=\arcsin(1)+2k\pi = \dfrac{\pi}{2}+2k\pi$$

therefore $x=\dfrac{\pi}{4}+k\pi,~k\in\mathbb{Z}$.

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  • $\begingroup$ More complicated way is to modify the product to sum $2\sin(x)\cos(x)=\sin(x+x)-sin(x-x)=\sin(2x)-\sin(0)$. Solutions are obviously the same. $\endgroup$
    – z100
    Commented Aug 5, 2019 at 17:13
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From $$\sin(2x)=2\sin x\cos x$$ $$\sin x\cos x = \frac{1}{2} \Leftrightarrow \sin(2x)=1$$

So the answer is $x=\frac{\pi}{4}$ in $0<x<\pi.$

The general solution is $$x=\frac{(4n\pi+\pi)}{4}, n\in\mathbb{Z}$$.

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There's a useful identity you can use.

$2\sin(x)\cos(x)=\sin(2x)$

Now, rewrite it as $\frac{\sin(2x)}{2}=\sin{x}\cos{x}$.

Swap $\frac{\sin(2x)}{2}$ with your LHS and you have:

$\frac{\sin(2x)}{2}=\frac{1}{2}$ as the following.

We are now left with $\sin(2x)=1$.

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Your solution is incorrect. You proved that $\sin x \cos x = \frac{1}{2}\implies\sin x =\pm\dfrac1{\sqrt2},$

but the reverse implication does not necessarily hold.

You need to look at which of your solutions satisfy the original equation. $135^o$ and $315^o$ do not.

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    $\begingroup$ In general, it's a good idea to see if the solutions you get satisfy the original equation; you might have introduced extraneous solutions or made an arithmetic mistake $\endgroup$ Commented Aug 5, 2019 at 17:03
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Your solution is not correct, because it assumes that $\cos x=\sqrt{1-\sin^2x}$. In fact, sometimes you indeed have $\cos x=\sqrt{1-\sin^2x}$, but sometimes you have $\cos x=-\sqrt{1-\sin^2x}$.

Note that\begin{align}\sin(x)\cos(x)=\frac12&\iff2\sin(x)\cos(x)=1\\&\iff\sin(2x)=1\\&\iff2x=\frac\pi2+2k\pi\text{ ($k\in\mathbb Z$)}\\&\iff x=\frac\pi4+k\pi\text{ ($k\in\mathbb Z$)}.\end{align}

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I am now going to add one more for fun and wish you can enjoy it. $$ \begin{aligned} &(\cos x+\sin x)^{2}=1+2 \sin x \cos x=2 \\ \Rightarrow & \cos x+\sin x=\pm \sqrt{2} \ldots(1) \\ &(\cos x-\sin x)^{2}=1-2 \sin x \cos x=0 \\ \Rightarrow & \cos x-\sin x=0 \quad \cdots(2) \end{aligned} $$ (1) $\pm$ (2) yields $$\quad \cos x=\sin x=\pm \frac{1}{\sqrt{2}}$$ $$ x=n \pi+\frac{\pi}{4} $$ where $n \in \mathbb{Z}.$

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