Question

Тригонометрические уравнения:cos^(-2) 2t-sin^(-2) 2t=8/3

Answer 1

$frac{1}{cos^22t} - frac{1}{sin^22t} =frac{8}{3} \ frac{sin^22t -cos^22t}{sin^22t *cos^22t}=frac{8}{3} \ npu cos2tneq 0 u sin2t neq 0:\ -3cos4t=8sin^22t *cos^22t\ -3cos4t=2sin^24t\ -3cos4t=2(1-cos^24t)\ 2cos^24t -3cos4t-2=0\ cos4t=y, y in [-1;1]\ 2y^2-3y-2=0\ y_1=2 notin [-1;1];\ y_2=-frac{1}{2} \ cos4t=-frac{1}{2}\ 4t= б frac{2pi}{3}+2pi k\ t= б frac{pi}{6}+frac{pi k}{2} ; npu tneq frac{pi k}{4}; k in Z$

Ответ: $б frac{pi}{6}+frac{pi k}{2}$

Answer 2

1/cos²(2t)-1/sin²(2t)=8/3;cos2t≠0;sin2t≠0
sin²2t-cos²2t=8/3*sin²2t*cos²2t
-cos4t=2/3*sin²4t
-3cos4t=2(1-cos²4t)
-3cos4t+2cos²4t-2=0
2cos²4t-3cos4t-2=0
cos4t=a
2a²-3a-2=9
D=9+16=25
a=(3±5)/4
a1=2;a2=-1/2
cos4t=-1/2
4t=±2π/3+2πk
t=±π/6+πk/2;k€Z