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求解 θ 的值
\theta =2\pi n_{2}+\frac{\pi }{3}\text{, }n_{2}\in \mathrm{Z}
\theta =2\pi n_{3}+\frac{5\pi }{3}\text{, }n_{3}\in \mathrm{Z}
\theta =\pi n_{1}\text{, }n_{1}\in \mathrm{Z}
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Trigonometry
5 道与此类似的题目:
4 \sin \theta \cos \theta = 2 \sin \theta
来自 Web 搜索的类似问题
Why do we lose solutions when we divide 2\sin\theta\cos\theta=\sin\theta by \sin\theta?
https://math.stackexchange.com/q/2806858
Whenever you divide both sides of an equation by something, you are assuming that the thing you're dividing by is nonzero, because dividing by 0 is not valid. So going from 2 \sin \theta \cos \theta = \sin \theta ...
Simple Trig Equations - Why is it Wrong to Cancel Trig Terms?
https://math.stackexchange.com/questions/1416619/simple-trig-equations-why-is-it-wrong-to-cancel-trig-terms
The two issues are: When cancelling a factor, note that this is only possible when the factor is not zero; but the factor may be zero in the solution to the problem. In this case \sin\theta = 0 ...
How do I simplify this expression further?
https://math.stackexchange.com/questions/472115/how-do-i-simplify-this-expression-further
You have \frac{1}{x^2+y^2} = 1+\left|\frac{2}{y^2/x^2+1}-1\right| = 1+\frac{|x^2-y^2|}{x^2+y^2}, or put differently 1 = x^2+y^2 + |x^2-y^2|. At this point solving by cases helps. Where |x|<|y| ...
Using the exponential form of a complex number and De Moivre's theorem
https://math.stackexchange.com/questions/1523448/using-the-exponential-form-of-a-complex-number-and-de-moivres-theorem
As you noted, \cos(2\theta+\pi/2) = \operatorname{Re}(e^{i(2\theta+\pi/2)}). e^{i(2\theta+\pi/2)}=(e^{i\theta})^2e^{i\pi/2}=(\cos\theta+i\sin\theta)^2(i) =i(\cos^2\theta+2i\cos\theta\sin\theta-\sin^2\theta) ...
Is the point of a shape with the greatest average ray length also the “centroid”?
https://math.stackexchange.com/questions/1588150/is-the-point-of-a-shape-with-the-greatest-average-ray-length-also-the-centroid
I have confused you with the definition of mean radius with the average ray length. When people think of average radius they think of "average squared minimized" or mean radius. What I am focusing on ...
Do we have to show it for both cases?
https://math.stackexchange.com/questions/1459841/do-we-have-to-show-it-for-both-cases
You can choose C=c+\pi, and then \sin{(\theta+C)}=-\sin{(\theta+c)}, so having the \pm there doesn't create any more solutions if you allow any c \in [0,2\pi).
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{ x } ^ { 2 } - 4 x - 5 = 0
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4 \sin \theta \cos \theta = 2 \sin \theta
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y = 3x + 4
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699 * 533
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\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
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\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
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\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
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\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
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\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}
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