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x=2\pi n_{1}+\frac{\pi }{2}
n_{1}\in \mathrm{Z}
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Trigonometry
\sin ( x ) + 2 = 3
Problemau tebyg o chwiliad gwe
How do you solve \displaystyle{\sin{{x}}}+{2}={3} ?
https://socratic.org/questions/how-do-you-solve-sinx-2-3
see below Explanation: \displaystyle{\sin{{x}}}+{2}={3} \displaystyle{\sin{{x}}}={3}-{2} \displaystyle{x}={{\sin}^{{-{{1}}}}{\left({1}\right)}}= \displaystyle{x}=\frac{\pi}{{2}}+{2}\pi{n}
How do you solve \displaystyle{3}{\sin{{x}}}+{2}={0} ?
https://socratic.org/questions/how-do-you-solve-3sinx-2-0
Nghi N. May 21, 2015 3sin x = -2 -> sin x = -2/3 2 answers: \displaystyle{\sin{{x}}}=-\frac{{2}}{{3}}\to{x}=-{41.81} deg \displaystyle{\sin{{x}}}=-\frac{{2}}{{3}}\to{x}=-{180}+{41.81}=-{138.19}
If \displaystyle{\sin{{x}}}={0.7} , how do you list all possible value of x if \displaystyle-{180}°{<}{x}{<}{180}° ?
https://socratic.org/questions/if-sin-x-0-7-how-do-you-list-all-possible-value-of-x-if-180-x-180
Nghi N. · Kevin B. Apr 17, 2015 \displaystyle{\sin{{x}}}={0.7} First, calculator gives \displaystyle{x}={44.42}{d}{e}{g} . The trig unit circle gives another arc \displaystyle{x}={180}-{44.42}={135}.{58}{d}{e}{g} ...
How do you graph \displaystyle-{2}{\sin{{x}}}+{2} ?
https://socratic.org/questions/how-do-you-graph-2sinx-2
\displaystyle-{2}{\sin{{x}}}+{2} can be graphed by starting with \displaystyle{\sin{{x}}} Explanation: The \displaystyle-{2} tells us two things: (a) the amplitude is 2, and (b) the ...
How do you solve \displaystyle{\sin{{x}}}+{3}={3} ?
https://socratic.org/questions/how-do-you-solve-sinx-3-3
\displaystyle{x}={0} Explanation: We have: \displaystyle{\sin{{\left({x}\right)}}}+{3}={3} First, let's subtract \displaystyle{3} from both sides of the equation: \displaystyle\Rightarrow{\sin{{\left({x}\right)}}}+{3}-{3}={3}-{3} ...
How do you write the cartesian equation for \displaystyle{r}={1}+{3}{\sin{{x}}} ?
https://socratic.org/questions/how-do-you-write-the-cartesian-equation-for-r-1-3sinx
As below. Explanation: \displaystyle{r}={1}+{3}{\sin{\theta}} \displaystyle{x}={r}{\cos{\theta}},{y}={r}{\sin{\theta}},{r}=\sqrt{{{x}^{{2}}+{y}^{{2}}}} \displaystyle\sqrt{{{x}^{{2}}+{y}^{{2}}}}={1}+{3}\cdot\frac{{y}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}} ...
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Problemau tebyg
\cos ( 3x + \pi ) = 0.5
\sin ( x ) = 1
\sin ( x ) - cos ( x ) = 0
\sin ( x ) + 2 = 3
{ \tan ( x ) } ^ {2} = 4
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