Как решить sin^2 33°+sin^2 33°-cos^60°

Question

Алгебра: Как решить sin^2 33°+sin^2 33°-cos^60°

MuxaBirko · Accepted Answer

1) sin15x*sin3x+cos7x*cos11x=0
1/2[cos(15x - 3x) - cos(15x + 3x)] - 1/2[cos(7x - 11x) + cos(7x + 11x)] = 0
1/2cos(12x) - 1/2cos(4x) = 0
cos(12x) - cos(4x) = 0
2*[sin( 12x + 4x)/2sin(4x - 12x)/2] = 0
sin(8x)sin(4x) = 0
1) sin(8x) = 0
8x = πn, n∈X
x1 = (πn)/8, n∈Z
2) sin(4x) = 0
4x = πk,k∈Z
x2 = (πk)/4, k∈Z

2) 1 - 3 sinx*cosx + cos^2x = 0
sin^2x + cos^2x - 3 sinx*cosx + cos^2x = 0
sin^2x + 2cos^2x - 3 sinx*cosx = 0 / cos^2x ≠ 0
tg^2x -3tgx + 2 = 0
1) tgx = 1
x1 = π/4 + πn, n∈Z
2) tgx = 2
x2 = arctg2 + πk, k∈Z

MOGZ ответил

Как решить
sin^2 33°+sin^2 33°-cos^60°