Question

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

Answer 1

$\frac{1}{cos^22t} - \frac{1}{sin^22t} =\frac{8}{3} \\ \frac{sin^22t -cos^22t}{sin^22t *cos^22t}=\frac{8}{3} \\ npu\ cos2t\neq 0\ u\ sin2t \neq 0:\\ -3cos4t=8sin^22t *cos^22t\\ -3cos4t=2sin^24t\\ -3cos4t=2(1-cos^24t)\\ 2cos^24t -3cos4t-2=0\\ cos4t=y,\ y \in [-1;1]\\ 2y^2-3y-2=0\\ y_1=2 \notin [-1;1];\\ y_2=-\frac{1}{2} \\ cos4t=-\frac{1}{2}\\ 4t= б\ \frac{2\pi}{3}+2\pi k\\ t= б\ \frac{\pi}{6}+\frac{\pi k}{2} ; npu\ t\neq \frac{\pi k}{4}; \ \ k \in Z$

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