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smile1411
@smile1411
July 2022
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5 sin в квадрате x - 3 sinx*cos x-2cos в квадрате x=0
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axe211
5sin^2x - 3sinxcosx - 2cosx = 0;
5sin^2x - 3sinxcox - cos^2x - sin^2x = 0 | : cos^2x;
5tg^2x - 3tgx -1 - tg^2x = 0 ;
4tg^2x - 3tgx - 1 = 0 ;
Делаем замену tgx = t ;
4t^2 - 3t - 1 =0
D= b^2 - 4ac = 9 - 4 *4 * (-1) = 25;
t1 = 3+5/8 = 1 ; t2 = 3-5/8 = -1/4;
tgx= 1 ; tgx = -1/4
x =p/4 + pk, k e z; x = arctg ( - 1/4) + pn, n e z;
ответ : x1 = p/4 +pk, k e z ; x2 = arctg ( - 1/4) + pn, n e z.
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Answers & Comments
5sin^2x - 3sinxcox - cos^2x - sin^2x = 0 | : cos^2x;
5tg^2x - 3tgx -1 - tg^2x = 0 ;
4tg^2x - 3tgx - 1 = 0 ;
Делаем замену tgx = t ;
4t^2 - 3t - 1 =0
D= b^2 - 4ac = 9 - 4 *4 * (-1) = 25;
t1 = 3+5/8 = 1 ; t2 = 3-5/8 = -1/4;
tgx= 1 ; tgx = -1/4
x =p/4 + pk, k e z; x = arctg ( - 1/4) + pn, n e z;
ответ : x1 = p/4 +pk, k e z ; x2 = arctg ( - 1/4) + pn, n e z.