Answer to Question #173531 in Algebra for Jon jay Mendoza

Question

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1.f(x)=2 sin (x-​π/​4)

2.f(x)=sin (x+π/​3)

3.f(x)=3 cos (x+π/​6)

4.f(x)=2 cos (x-2π)

5.f(x)=tan x

Answer 1

$\begin{array}{l} \text { 1) } f(x)=2 \sin \left(x-\frac{\pi}{4}\right)=0 \\ \sin \left(x-\frac{\pi}{4}\right)=0, \quad y=x-\frac{\pi}{4} \\ \sin y=0 \\ k \in Z \quad y=\pi k \quad x-\frac{\pi}{4}=\pi k \\ x=\frac{\pi}{4}+\pi k \end{array}$

$\begin{array}{l} \text { d) } 4) f(x)=2 \cos (x-2 \pi)=0 \\ \cos (x-2 \pi)=0 \\ \cos y=0 \quad y=x-2 \pi \\ y=\frac{\pi}{2}+\pi k, \quad x=\frac{\pi}{2}+2 \pi+\pi k \quad k \in z \\ x=\frac{5 \pi}{2}+\pi k \Rightarrow x=\frac{\pi}{2}+\pi k \end{array}$ $\begin{array}{l} \text { 3) } \begin{aligned} f(x)=3 \cos (x+\pi / 6)=0 \\ \cos (x+\pi / 6)=0 \end{aligned} \\ \text { Ces } \begin{array}{rl} y=x+\pi / 6 \quad \cos y=0 \\ y=\frac{\pi}{2}+\pi k \quad x=\frac{\pi}{2}-\frac{\pi}{6}+\pi k & k \in Z \\ x=\frac{\pi}{3}+\pi k \end{array} \end{array}$ $\text { 5) } \begin{aligned} f(x) &=\tan x=0 \\ x &=\frac{\pi}{2}+\pi k \quad k \in Z \end{aligned}$ $\text { 2)} \begin{array}{rl} f(x)=\sin (x+\pi / 3)=0 \\ \sin y=0, & y=x+\pi / 3 \\ y=10 \pi k & x+\pi / 3=\pi k \quad k \in Z \\ x & =-\pi / 3+\pi k \end{array}$