2sin(Квадрат)x+7cosx-5=0

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[latex]2sin^{2}(x) + 7cos(x)-5 = 0\ 2(1-cos^{2}(x))+ 7cos(x)-5 = 0\ 2cos^{2}(x)-7cos(x)+3=0\ cos(x)_{1,2} = frac{7бsqrt{25}}{4} = frac{7б5}{4}\ \ t.k. |cos(x)|leq 1, to \ => cos(x) = frac{1}{2}\ x = бfrac{pi}{3} + 2pi n, n - zeloe [/latex]