Question

2cos^2(x/3)+3sin(x/3)-3=0

Answer 1

Task:

Solve

$$2 \cos^ {2} \left(\frac {\mathrm {X}}{3}\right) + 3 \sin \left(\frac {\mathrm {X}}{3}\right) - 3 = 0$$

Solution:

$$2 \cos^ {2} \left(\frac {\mathrm {X}}{3}\right) + 3 \sin \left(\frac {\mathrm {X}}{3}\right) - 3 = 0$$ $$2 \left(1 - \sin^ {2} \left(\frac {\mathrm {X}}{3}\right)\right) + 3 \sin \left(\frac {\mathrm {X}}{3}\right) - 3 = 0$$ $$- 2 \sin^ {2} \left(\frac {\mathrm {X}}{3}\right) + 3 \sin \left(\frac {\mathrm {X}}{3}\right) - 1 = 0$$ $$2 \sin^ {2} \left(\frac {\mathrm {X}}{3}\right) - 3 \sin \left(\frac {\mathrm {X}}{3}\right) + 1 = 0$$ $$D = 9 - 2 * 4 = 1$$ $$\sin \left(\frac {x}{3}\right) = \frac {3 \pm 1}{4}$$ $$\left[ \begin{array}{l} \sin \left(\frac {\mathrm {x}}{3}\right) = 1 \\ \sin \left(\frac {\mathrm {x}}{3}\right) = \frac {1}{2} \end{array} \right. = > \left[ \begin{array}{l} \frac {\mathrm {x}}{3} = \frac {\pi}{2} + 2 \pi \mathrm {n}, \mathrm {n} \in \mathrm {Z} \\ \frac {\mathrm {x}}{3} = \frac {\pi}{6} + 2 \pi \mathrm {k}, \mathrm {k} \in \mathrm {Z} \\ \frac {\mathrm {x}}{3} = \frac {5 \pi}{6} + 2 \pi \mathrm {m}, \mathrm {m} \in \mathrm {Z} \end{array} \right. = > \left[ \begin{array}{l} \mathrm {x} = \frac {3 \pi}{2} + 6 \pi \mathrm {n}, \mathrm {n} \in \mathrm {Z} \\ \mathrm {x} = \frac {\pi}{2} + 6 \pi \mathrm {k}, \mathrm {k} \in \mathrm {Z} \\ \mathrm {x} = \frac {5 \pi}{2} + 6 \pi \mathrm {m}, \mathrm {m} \in \mathrm {Z} \end{array} \right]$$

Answer: $\frac{\pi}{2} + 6\pi \mathrm{k}, \quad \frac{3\pi}{2} + 6\pi \mathrm{n}, \quad \frac{5\pi}{2} + 6\pi \mathrm{m}; \quad \mathrm{n}, \mathrm{m}, \mathrm{k} \in \mathrm{Z}$