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Cari nilai x
x=\pi n_{1}+\frac{\pi }{4}
n_{1}\in \mathrm{Z}
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Kedua Sisi Grafik dalam 2D
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Trigonometry
\sin ( x ) = \cos ( x )
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https://math.stackexchange.com/questions/1476944/how-to-solve-equations-like-2-sinx-cosx/1476973
One way can be using tan\frac x2=t so sin x=\frac{2t}{1+t^2} and cos x=\frac{1-t^2}{1+t^2}. Here 2sin x= cos x implies t^2+4t-1=0 from wich tan \frac x2=2\pm\sqrt{5}.Hence the answer of ...
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https://socratic.org/questions/how-do-you-show-that-the-equation-1-sinx-cosx-is-not-an-identity
Bdub Nov 12, 2016 Pick a value for x like \displaystyle\frac{\pi}{{3}} and plug it in to both side to show that they don't equal each other and therefore not an identity
How do you solve \displaystyle{1}+{\sin{{\left({x}\right)}}}={\cos{{\left({x}\right)}}} ?
https://socratic.org/questions/how-do-you-solve-1-sin-x-cos-x
\displaystyle{x}={0} Explanation: \displaystyle{1}+{\sin{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}{\quad\text{or}\quad}{\cos{{x}}}-{\sin{{x}}}={1} . Squaring both sides we get \displaystyle{\left({\cos{{x}}}-{\sin{{x}}}\right)}^{{2}}={1}{\quad\text{or}\quad}{{\cos}^{{2}}{x}}+{{\sin}^{{2}}{x}}-{2}{\sin{{x}}}{\cos{{x}}}={1}{\quad\text{or}\quad}{1}-{\sin{{2}}}{x}={1}{\quad\text{or}\quad}{\sin{{2}}}{x}={0}={\sin{{0}}}; ...
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As @Nicholas Stull hinted, you lost solutions by not making sure that you were not dividing by zero. As @Winther pointed out, you can avoid this error by factoring. As @Nicholas Stull pointed out, ...
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https://math.stackexchange.com/q/2454114
Apropos "deeper way": 1) f(x) = f(-x), even fct. Examples: y=x^2, y=cos(x) f'(x) = -f'(-x), chain rule, odd fct. 2) f(x)=-f(-x), odd fct. Examples: y=x^3, y=sin(x). f'(x) = f'(-x), ...
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https://math.stackexchange.com/q/2212333
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\tan ( x )
\sec ( x )
\sin ( x ) = \cos ( x )
\cot ( x )
\cos ( x )
\csc ( x )
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