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Help me with this, I'm stuck:

If $\operatorname{tg}x-\operatorname{ctg}x=-0.75$ with condition $\dfrac{3 \pi}{2}<x<\dfrac{7\pi}{4}$, calculate $\cos2x$.

What I did:

$$\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}=-0.75$$

$$\frac{\sin^2x-\cos^2x}{\sin x \cos x}=-0.75$$

$$\frac{2\cos2x}{\sin2x}=0.75$$

$$\frac{\cos2x}{\sin2x}=3/8$$

$$\operatorname{tg}2x=\frac{8}{3}$$

$$x=0.6060128283$$

I don't know now how to fulfill the condition, because the angle isn't between the marks, even if I flipped it -- if $x=-0.6060128283$, that's (around $5.68$) still not in the condition - from around $4.7$ to around $5.5$.

The result in my book is $\dfrac{-3}{\sqrt{73}}$, which is a negative from a result I'd get, which makes no sense since cosinus should be positive... (EDIT: wrong, it makes sense)

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  • $\begingroup$ $\cos x$ is positive for $\frac{3\pi}{2} < x < \frac{7\pi}{4}$, but $\cos (2x)$ is negative then. $\endgroup$ – Daniel Fischer May 17 '14 at 12:20
  • $\begingroup$ Ok, i realized that when I typed this question... What about the condition? Ok, I'm just stupid. $\endgroup$ – guest May 17 '14 at 12:24
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$$\frac{3\pi}{2}<x<\frac{7\pi}{4}\Rightarrow 3 \pi<2x<\frac{7\pi}{2}\Rightarrow \pi<2x< \frac{3\pi}{2}$$ (Handle with care)

So, $\cos(2x)$ is negative....

I do not have complete solution with me at this time but i am sure that answer given in the book is correct....

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