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$$\begin{align}\cos^2(x)&=1-\sin(x)\\ 1-\sin^2(x)&=1-\sin(x)\\ (1-\sin x)(1+\sin x)&= 1-\sin(x) \end{align}$$ divide both sides by $1 - \sin(x)$

End up with $1 + \sin(x)$

The answer is supposed to be in radians between $0$ and $2 \pi$.

So I get $1+\sin(x)=0$

$$\sin(x)=-1 = -90\text{ degrees } = -\pi/2 \text{ or }3\pi/2$$

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    $\begingroup$ What if $1-\sin{x}=0$? $\endgroup$ Jul 8 '15 at 1:05
  • $\begingroup$ You got $-\pi/2$ which is not in the interval, and you lost $\sin x=1$ ($x=\pi/2$). $\endgroup$ Jul 8 '15 at 1:06
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    $\begingroup$ Note that "$\sin(x)=-1=-90\text{ degrees}\dots$" is really bad form. $-1$ is not equal to $-90^o$. What you mean is "$\sin x=-1$ implies $x=-90^o$. $\endgroup$ Jul 8 '15 at 1:08
  • $\begingroup$ No. You get $1 - \sin x = 1$, $\endgroup$ Jul 8 '15 at 6:36
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You are almost correct: at the step: $1 - \sin^2 x = 1 - \sin x$ It would be easier to do the following: $$\sin^2 x - \sin x = \sin{x} (\sin{x} - 1) = 0$$

Thus, we would have $\sin{x} = 0$ and $\sin{x} = 1$

From here it follows that your solutions are, in the domain $[0, 2\pi)$,

$$0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$$

Unfortunately, you potentially divided by 0 without accounting for it: you have to consider when $\sin{x} = 1$; Also, pay mind to your domain.

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  • $\begingroup$ I see my mistake now. Thanks! $\endgroup$
    – user253112
    Jul 8 '15 at 1:10
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    $\begingroup$ @user253112 In general, don't divide, factorise! $\endgroup$
    – Deepak
    Jul 8 '15 at 1:23
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If you directly divide by $1-\sin x$, you miss the solution $x=\frac{\pi}{2}$, since it makes both side zero. After noting this, you should say $1+\sin x=1$ not $0$. This gives $\sin x=0$, which gives you another solution $x=\pi$.

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When you divide by $1-\sin x$ you should get $1+\sin x = 1$, not $0$.

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There are two correct ways to proceed from $$(1-\sin x)(1+\sin x) = 1-\sin(x)$$

first way

This will be true if

$1 - \sin x = 0$

$x \in \{90^{\circ}, 270^{\circ}\}$

If not, then we can divide both sides by $1 - \sin x$, getting

$1 + \sin x = 1$

$\sin x = 0$

$x \in \{0^{\circ}, 180^{\circ}\}$

second way

$$(1-\sin x)(1+\sin x) = 1-\sin(x)$$

$$(1-\sin x)(1+\sin x) - (1-\sin x) = 0$$

$$(1-\sin x)(1+\sin x - 1) = 0$$

$$(1-\sin x)\sin x = 0$$

$$1 - \sin x = 0 \; \text {or} \; \sin x = 0\ldots$$

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Hey there you are partially correct , one thing you missed in your solution Follow my solution $\cos^2x=1-\sin x$

$\Longrightarrow (1-\sin x)(1+\sin x)=1-\sin x$

From this equation we have

$1- \sin x=0$ and $1+ \sin x=0$

$\sin x = 1$ and $\sin x = -1$

$\sin x= \sin(\frac\pi2)$ and $\sin x= \sin(-\frac\pi2)$ We have if $\sin x= \sin\alpha$ then $x= n\pi+(-1)^n\alpha$

Use this result for both equations and find solution.

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