1) 5cos в квадрате x +6sinx-6=0 2)2tg в квадрате x+tgx-2=0 3)sinx-3cosx=0 4)корень из 3*sinx*cosx+3cosв квадрате x=0

1) 5cos^2 x + 6sinx - 6 = 0;

5(1 - sin^2x) + 6sinx - 6 = 0;

5 - 5sin^2x + 6sinx - 6 = 0;

-5sin^2x + 6sinx - 1 = 0;

5sin^2x - 6sinx + 1 = 0;

5t^2 - 6t + 1 = 0;

D = 36 - 4 * 5 = 16;

t = (6 +- 4)/10;

t1 = 1; t2 = 0,2;

sinx = 1 => x = pi/2 + 2pin, n ∈ Z;

sinx = 0,2 => x = (-1)^narcsin0,2 + pin, n ∈ Z;

2) 2tg^2 x + tgx - 2 = 0;

2t + t - 2 = 0;

D = 1 + 4 * 2 * 2 = 17;

t = (-1 +- sqrt17)/4;

tgx = (-1 + sqrt17)/2 => x = arctg(-1 + sqrt17)/2 + pin, n ∈ Z;

tgx = (-1 - sqrt17)/2 => x = arctg(-1 - sqrt17)/2 + pin, n ∈ Z;

3) sinx - 3cosx = 0;

tgx - 3 = 0;

tgx = 3;

x = arctg3 + pin, n ∈ Z;

4) sqrt3 * sinx * cosx + 3cos^2 x = 0;

cosx(sqrt3sinx + 3cosx) = 0;

cosx = 0 => x = pi/2 + pin,  n ∈ Z;

sqrt3sinx + 3cosx = 0;

sqrt3tgx + 3 = 0;

sqrt3tgx = -3;

tgx = -3/sqrt3;

tgx = -sqrt3 => x = 2pi/3 + pin, n ∈ Z;

5) sin^2 (3x/4) -sqrt2/2 = sinx - cos^2(3x/4)+1;

1 - sqrt2/2 = sinx + 1;

sinx = -sqrt2/2 => x = (-1)^n5pi/4 + pin, n ∈ Z;

6) 2cos^2 x/2+ sqrt3cosx/2 = 0;

Пусть х/2 = a;

2cos^2a + sqrt3cosa = 0;

cosa(2cosa + sqrt3) = 0;

cosa = 0 => a = pi/2 + pin => x = pi + 2pin, n ∈ Z;

2cosa + sqrt3 = 0;

2cosa = -sqrt3;

cosa = -sqrt3/2 => a = +-5pi/6 + 2pin => x = +-5pi/3 + 4pin, n ∈ Z;

7) 2sin^2 2x - 5sin2x * cos2x + 2cos^2 2x = 0;

2tg^2 2x - 5tg2x  + 2 = 0;

2t - 5t + 2 = 0;

D = 25 - 4 * 4 = 9;

t = (5 +- 3)/4;

t1 = 1/2, t2 = 2;

tg2x = 1/2 => 2x = arctg1/2 + pin => x = (arctg1/2)/2 + pi/2n,  n ∈ Z;

tg2x = 2  => 2x = arctg2 + pin => x = (arctg2)/2 + pin/2,  n ∈ Z;

8) 5sin^2 x - 14sinx * cosx - 3cos^2 x = 2;

5sin^2x - 14sinxcosx - 3cos^2x = 2sin^2x + 2cos^2x;

5sin^2x - 14sinxcosx - 3cos^2x - 2sin^2x - 2cos^2x = 0;

3sin^2x - 14sinxcosx - 5cos^2x = 0;

3tg^2x - 14tgx - 5 = 0;

3t^2 - 14t - 5 = 0;

D = 14^2 + 4 * 3 * 5 = 256;

t = (14 +- 16)/ 6;

t1 = 5, t2 = -1/3;

x = arctg5 + pin,  n ∈ Z;

x = arctg(-1/3) +pin,  n ∈ Z.