6 (sinx+cosx) - 2sinx•cosx+6=0

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14 марта, 05:43

6 (sinx+cosx) - 2sinx•cosx+6=0

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Ионка · Answer 1

6 * (sin (x) + cos (x)) - 2*sin (x) * cos (x) + 6=0; 6 * (sin (x) + cos (x)) - 2*sin (x) * cos (x) + 7 - (sin (x)) ^2 - (cos (x)) ^2=0;

6 * (sin (x) + cos (x)) - (sin (x) + cos (x)) ^2+7=0; (sin (x) + cos (x)) ^2-6 * (sin (x) + cos (x)) - 7=0; 1). (sin (x) + cos (x)) = - 1;

1+sin (x) + cos (x) = 0; 2 * (cos (0,5*x)) ^2+2*sin (0,5*x) * cos (0,5*x) = 0; cos (0,5*x1) = 0; tg (0,5*x2) = - 1