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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
"\\begin{array}{l}\n\\text { 1) } f(x)=2 \\sin \\left(x-\\frac{\\pi}{4}\\right)=0 \\\\\n\\sin \\left(x-\\frac{\\pi}{4}\\right)=0, \\quad y=x-\\frac{\\pi}{4} \\\\\n\\sin y=0 \\\\\nk \\in Z \\quad y=\\pi k \\quad x-\\frac{\\pi}{4}=\\pi k \\\\\nx=\\frac{\\pi}{4}+\\pi k\n\\end{array}"
"\\begin{array}{l}\n\\text { d) } 4) f(x)=2 \\cos (x-2 \\pi)=0 \\\\\n\\cos (x-2 \\pi)=0 \\\\\n\\cos y=0 \\quad y=x-2 \\pi \\\\\ny=\\frac{\\pi}{2}+\\pi k, \\quad x=\\frac{\\pi}{2}+2 \\pi+\\pi k \\quad k \\in z \\\\\nx=\\frac{5 \\pi}{2}+\\pi k \\Rightarrow x=\\frac{\\pi}{2}+\\pi k\n\\end{array}" "\\begin{array}{l}\n\\text { 3) } \\begin{aligned}\nf(x)=3 \\cos (x+\\pi \/ 6)=0 \\\\\n\\cos (x+\\pi \/ 6)=0\n\\end{aligned} \\\\\n\\text { Ces } \\begin{array}{rl}\ny=x+\\pi \/ 6 \\quad \\cos y=0 \\\\\ny=\\frac{\\pi}{2}+\\pi k \\quad x=\\frac{\\pi}{2}-\\frac{\\pi}{6}+\\pi k & k \\in Z \\\\\nx=\\frac{\\pi}{3}+\\pi k\n\\end{array}\n\\end{array}" "\\text { 5) } \\begin{aligned}\nf(x) &=\\tan x=0 \\\\\nx &=\\frac{\\pi}{2}+\\pi k \\quad k \\in Z\n\\end{aligned}" "\\text { 2)} \\begin{array}{rl}\nf(x)=\\sin (x+\\pi \/ 3)=0 \\\\\n\\sin y=0, & y=x+\\pi \/ 3 \\\\\ny=10 \\pi k & x+\\pi \/ 3=\\pi k \\quad k \\in Z \\\\\nx & =-\\pi \/ 3+\\pi k\n\\end{array}"
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