Assorted trigonometric equations and inequalities

Text

Solve the following equations and inequalities:

(a) $3-2\cos\alpha=2$

(b) $2-3\sin\alpha\le 4$

(c) $3\tan^2\alpha-\tan\alpha=1$

(d) $1-3\sin(3\alpha+\pi)=0$

(e) $\cos\alpha\le\dfrac{\sqrt3}{3}$

(f) $\cos^2\alpha+\dfrac12\cos\alpha=0$

Solution

(a) $\cos\alpha=\dfrac12 \;\Longrightarrow\; \alpha=\pm\dfrac\pi3+2k\pi$.

(b) The inequality is equivalent to $\sin\alpha\ge-\dfrac23$: $\alpha\in\left[\arcsin\!\left(-\frac23\right)+2k\pi,\ \ \pi-\arcsin\!\left(-\frac23\right)+2k\pi\right],\qquad \arcsin\!\left(-\frac23\right)\approx-41.8°.$

(c) Moving everything to one side, $3\tan^2\alpha-\tan\alpha-1=0$, so $\tan\alpha=\frac{1\pm\sqrt{13}}{6}\approx 0.768 \ \text{ or } \ -0.434,$ hence $\alpha\approx37.5°+k\cdot180°$ or $\alpha\approx-23.5°+k\cdot180°$.

(d) Since $\sin(3\alpha+\pi)=-\sin3\alpha$, the equation becomes $\sin3\alpha=-\dfrac13$, so $3\alpha=\arcsin\!\left(-\frac13\right)+2k\pi \ \text{ or } \ 3\alpha=\pi-\arcsin\!\left(-\frac13\right)+2k\pi,$ and finally one divides by $3$ to obtain $\alpha$.

(e) Since $\dfrac{\sqrt3}{3}\approx0.577$: $\alpha\in\left[\arccos\frac{\sqrt3}{3},\ \ 2\pi-\arccos\frac{\sqrt3}{3}\right]\pmod{2\pi},\qquad \arccos\frac{\sqrt3}{3}\approx54.7°.$

(f) Factoring, $\cos\alpha\left(\cos\alpha+\dfrac12\right)=0$, so $\cos\alpha=0$ (i.e. $\alpha=\dfrac\pi2+k\pi$) or $\cos\alpha=-\dfrac12$ (i.e. $\alpha=\pm\dfrac{2\pi}3+2k\pi$).

$\boxed{\alpha=\pm\frac\pi3+2k\pi;\ \ \sin\alpha\ge-\frac23;\ \ \tan\alpha=\frac{1\pm\sqrt{13}}{6};\ \ \sin3\alpha=-\frac13;\ \ \cos\alpha\le\frac{\sqrt3}{3};\ \ \cos\alpha\in\left\{0,-\frac12\right\}}$