Question

Just the answer please.
1: http://imgur.com/pSoypwa
2: http://imgur.com/jiEwTJa
3: http://imgur.com/OFi4bf7

Answer 1

Answer on Question #58926 – Math – Trigonometry

Question

1. Solve on the interval $[0, 2\pi)$:

$$2 \sin^2 x - 3 \sin x + 1 = 0$$

Solution

$$2 \sin^2 x - 3 \sin x + 1 = 0, \quad 0 \leq x < 2\pi. \quad \text{Let } t = \sin(x), \quad -1 \leq t \leq 1.$$

Then $2t^2 - 3t + 1 = 0$, $D = 3^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1$, $t = \frac{3 \pm \sqrt{D}}{2 \cdot 2} = \frac{3 \pm 1}{4} = \frac{3 + 1}{4}$; $\frac{3 - 1}{4} = 1$; $\frac{1}{2}$, hence $\sin(x) = 1$ or $\sin(x) = \frac{1}{2}$ and finally obtain $x = \frac{\pi}{2}$, $x = \frac{\pi}{6}$, $x = \frac{5\pi}{6}$.

Answer: $x = \frac{\pi}{2}$, $x = \frac{\pi}{6}$, $x = \frac{5\pi}{6}$.

Question

2. Solve

$$tgx(tgx - 1) = 0$$

Solution

$$tgx(tgx - 1) = 0 \Rightarrow tgx = 0 \quad \text{or} \quad tgx = 1 \Rightarrow x = \pm \pi n \quad \text{or} \quad x = \frac{\pi}{4} \pm \pi n, \quad \text{where } n \text{ is integer}.$$

Answer: $x = \pm \pi n$, $x = \frac{\pi}{4} \pm \pi n$.

Question

3. Solve on the interval $[0, 2\pi)$:

$$1 - \cos \theta = \frac{1}{2}.$$

Solution

$$1 - \cos \theta = \frac{1}{2}, \quad 0 \leq \theta < 2\pi \Rightarrow \cos \theta = \frac{1}{2} \Rightarrow \theta = \frac{2\pi}{3}, \quad \theta = \frac{4\pi}{3} \quad \text{on the interval } [0, 2\pi).$$

Answer: $\theta = \frac{2\pi}{3}$, $\theta = \frac{4\pi}{3}$.

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