- AFinite solution
- BInfinite solution
- COne solution
- ✓No solution
==> $\frac{3}{5}\cos x + \frac{4}{5}\sin x = \frac{6}{5}$
==> $\cos (x - \theta ) = \frac{6}{5}$,
$[{\rm{where\,\, }}\theta = {\cos ^{ - 1}}(3/5)]$
So, that equation has no solution.
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Statement $I$ : For any two non-zero complex numbers $\mathrm{z}_1, \mathrm{z}_2$
$\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$ and
Statement $II$ : If $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are three distinct complex numbers and a, b, c are three positive real numbers such that $\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}$, then
$\frac{\mathrm{a}^2}{\mathrm{y}-\mathrm{z}}+\frac{\mathrm{b}^2}{\mathrm{z}-\mathrm{x}}+\frac{\mathrm{c}^2}{\mathrm{x}-\mathrm{y}}=1$
Between the above two statements,