Solve for E (complex solution)
E=\frac{2\cos(2x)-\sin(x)-\sin(3x)+2}{2}
\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}
Solve for E
E=-2\left(\sin(x)-1\right)\left(\cos(x)\right)^{2}
\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}
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\frac{E}{-\left(\cos(x)\right)^{2}}-2\sin(x)=-2
Subtract 2 from both sides. Anything subtracted from zero gives its negation.
\frac{E}{-\left(\cos(x)\right)^{2}}=-2+2\sin(x)
Add 2\sin(x) to both sides.
\left(-\frac{1}{\left(\cos(x)\right)^{2}}\right)E=2\sin(x)-2
The equation is in standard form.
\frac{\left(-\frac{1}{\left(\cos(x)\right)^{2}}\right)E}{-\frac{1}{\left(\cos(x)\right)^{2}}}=\frac{2\left(\sin(x)-1\right)}{-\frac{1}{\left(\cos(x)\right)^{2}}}
Divide both sides by -\left(\cos(x)\right)^{-2}.
E=\frac{2\left(\sin(x)-1\right)}{-\frac{1}{\left(\cos(x)\right)^{2}}}
Dividing by -\left(\cos(x)\right)^{-2} undoes the multiplication by -\left(\cos(x)\right)^{-2}.
E=-2\left(\sin(x)-1\right)\left(\cos(x)\right)^{2}
Divide 2\left(-1+\sin(x)\right) by -\left(\cos(x)\right)^{-2}.
\frac{E}{-\left(\cos(x)\right)^{2}}-2\sin(x)=-2
Subtract 2 from both sides. Anything subtracted from zero gives its negation.
\frac{E}{-\left(\cos(x)\right)^{2}}=-2+2\sin(x)
Add 2\sin(x) to both sides.
\left(-\frac{1}{\left(\cos(x)\right)^{2}}\right)E=2\sin(x)-2
The equation is in standard form.
\frac{\left(-\frac{1}{\left(\cos(x)\right)^{2}}\right)E}{-\frac{1}{\left(\cos(x)\right)^{2}}}=\frac{2\left(\sin(x)-1\right)}{-\frac{1}{\left(\cos(x)\right)^{2}}}
Divide both sides by -\left(\cos(x)\right)^{-2}.
E=\frac{2\left(\sin(x)-1\right)}{-\frac{1}{\left(\cos(x)\right)^{2}}}
Dividing by -\left(\cos(x)\right)^{-2} undoes the multiplication by -\left(\cos(x)\right)^{-2}.
E=-2\left(\sin(x)-1\right)\left(\cos(x)\right)^{2}
Divide 2\left(-1+\sin(x)\right) by -\left(\cos(x)\right)^{-2}.
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