Cos²x+sinxcosx=1 [-π;π]
sinxcosx-(1-cos²x)=0
sinxcosx-sin²x=0
sinx(cosx-sinx)=0
sinx=0 или cosx-sinx=0 |:cosx≠0
x=πn, n∈Z 1-tgx=0
tgx=1
x=π/4+πn, n∈Z
x∈[-π;π]
x₁=-π
x₂=-π+π/4=-3π/4
x₃=0
x₄=π/4
x₅=π
(x₁+x₂+x₃+x₄+x₅):5=(-π -3π/4 +0 +π/4 +π):5=(-π/2):5=-π/10
Ответ: -π/10
![\frac{ \sqrt{3}sin2x }{1+cos2x}=1\; \; \; \; \; [- \pi ;2 \pi ]\\\\ \frac{ \sqrt{3}*2sinxcosx }{sin^2x+cos^2x+cos^2x-sin^2x}=1\\\\ \frac{2 \sqrt{3}sinxcosx }{2cos^2x}=1\\\\ \sqrt{3}tgx=1\\\\tgx= \frac{\sqrt{3}}{3} \\\\x_1=-5 \pi /6\; \; \; \; x_2= \pi /6\; \; \; \; x_3=7 \pi /6](https://tex.z-dn.net/?f=+%5Cfrac%7B+%5Csqrt%7B3%7Dsin2x+%7D%7B1%2Bcos2x%7D%3D1%5C%3B+%5C%3B+%5C%3B+%5C%3B+%5C%3B+%5B-+%5Cpi+%3B2+%5Cpi+%5D%5C%5C%5C%5C+%5Cfrac%7B+%5Csqrt%7B3%7D%2A2sinxcosx+%7D%7Bsin%5E2x%2Bcos%5E2x%2Bcos%5E2x-sin%5E2x%7D%3D1%5C%5C%5C%5C+%5Cfrac%7B2+%5Csqrt%7B3%7Dsinxcosx+%7D%7B2cos%5E2x%7D%3D1%5C%5C%5C%5C+%5Csqrt%7B3%7Dtgx%3D1%5C%5C%5C%5Ctgx%3D+%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D+%5C%5C%5C%5Cx_1%3D-5+%5Cpi+%2F6%5C%3B+%5C%3B+%5C%3B+%5C%3B+x_2%3D+%5Cpi+%2F6%5C%3B+%5C%3B+%5C%3B+%5C%3B+x_3%3D7+%5Cpi+%2F6++++ "\frac{ \sqrt{3}sin2x }{1+cos2x}=1\; \; \; \; \; [- \pi ;2 \pi ]\\\\ \frac{ \sqrt{3}*2sinxcosx }{sin^2x+cos^2x+cos^2x-sin^2x}=1\\\\ \frac{2 \sqrt{3}sinxcosx }{2cos^2x}=1\\\\ \sqrt{3}tgx=1\\\\tgx= \frac{\sqrt{3}}{3} \\\\x_1=-5 \pi /6\; \; \; \; x_2= \pi /6\; \; \; \; x_3=7 \pi /6")
Ответ: 3 корня