Question

Just the answers please.
1: http://imgur.com/CIY591W
2: http://imgur.com/UfOVG9N
3: http://imgur.com/CKdl1XK

Answer 1

Answer on Question #59340 – Math – Trigonometry

Question

The value $\frac{\pi}{24}$ is a solution for the equation $4\cos^2(4x) - 3 = 0$.

False

True

Solution

$$4 \cos^2 \left(4 \cdot \frac{\pi}{24}\right) - 3 = 4 \cos^2 \left(\frac{\pi}{6}\right) - 3 = 4 \left(\frac{\sqrt{3}}{2}\right)^2 - 3 = 3 - 3 = 0.$$

Answer: True.

Question

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

$$(\cos x + 1)(2 \cos^2 x - 3 \cos x - 2) = 0$$ $$x = 2\pi, \quad x = \frac{\pi}{2}, \quad x = \frac{\pi}{3}$$ $$x = \frac{\pi}{6}, \quad x = \frac{7\pi}{6}$$ $$x = \pi, \quad x = \frac{2\pi}{3}, \quad x = \frac{4\pi}{3}$$ $$x = 2\pi, \quad x = \frac{\pi}{2}, \quad x = \frac{5\pi}{4}$$

Solution

$$(\cos x + 1)(2 \cos^2 x - 3 \cos x - 2) = 0, \quad 0 \leq x < 2\pi;$$ $$\cos x + 1 = 0, \quad 0 \leq x < 2\pi, \quad \text{or} \quad 2 \cos^2 x - 3 \cos x - 2 = 0, \quad 0 \leq x < 2\pi;$$ $$\cos x + 1 = 0, \quad 0 \leq x < 2\pi;$$ $$\cos x = -1, \quad 0 \leq x < 2\pi ;$$ $$x = \pi ;$$ $$2 \cos^2 x - 3 \cos x - 2 = 0, \quad 0 \leq x < 2\pi ;$$ $$D = (-3)^2 - 4 \cdot 2 \cdot (-2) = 9 + 16 = 25;$$ $$\cos x = \frac{3 + 5}{2 \cdot 2} = \frac{8}{4} = 2 \quad \text{or} \quad \cos x = \frac{3 - 5}{2} = -1, \quad 0 \leq x < 2\pi ;$$

$\cos x = 2$ does not have real solutions on $0 \leq x < 2\pi$;

$\cos x = -\frac{1}{2}$ has solutions $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$ on $0 \leq x < 2\pi$.

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

Question

Evaluate $\cot\left(\cos^{-1}\left(-\frac{15}{17}\right)\right)$. Enter your answer as a fraction using the slash bar $[\text{?}]$.

Solution

$$\begin{array}{l} \cot(\cos^{-1}(x)) = x / \sqrt{1 - x^2}; \\ \cot\left(\cos^{-1}\left(-\frac{15}{17}\right)\right) = -\frac{15}{17} / \sqrt{1 - \left(-\frac{15}{17}\right)^2} = -\frac{15}{17} / \frac{8}{17} = -\frac{15}{8} = -1.875. \end{array}$$

Answer: $-15 / 8$.

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