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

Answer 3

To solve the trigonometric equation cos^(-2) 2t - sin^(-2) 2t = 8/3, we can use some trigonometric identities to simplify it and then solve for the value of t. Let's go step by step:

Step 1: Use the Pythagorean identity: sin^2 θ + cos^2 θ = 1

We can rewrite the equation as follows:

(1/cos^2 2t) - (1/sin^2 2t) = 8/3

Step 2: Find a common denominator for the fractions. The common denominator will be sin^2 2t * cos^2 2t:

(sin^2 2t - cos^2 2t) / (sin^2 2t * cos^2 2t) = 8/3

Step 3: Use the difference of squares identity: sin^2 θ - cos^2 θ = -cos(2θ)

(-cos(2t)) / (sin^2 2t * cos^2 2t) = 8/3

Step 4: Now, we can get rid of the fractions by multiplying both sides by sin^2 2t * cos^2 2t:

-1 = (8/3) * (sin^2 2t * cos^2 2t)

Step 5: Divide both sides by (8/3) to solve for sin^2 2t * cos^2 2t:

sin^2 2t * cos^2 2t = -3/8

Step 6: Apply another trigonometric identity: sin(2θ) = 2 * sin θ * cos θ

sin^2 2t * cos^2 2t = (2 * sin 2t * cos 2t)^2

(2 * sin 2t * cos 2t)^2 = -3/8

Step 7: Take the square root of both sides:

2 * sin 2t * cos 2t = ±√(-3/8)

Step 8: Solve for sin 2t * cos 2t:

sin 2t * cos 2t = ±√(-3/8) / 2

sin 2t * cos 2t = ±√(-3) / 4√2

Since the value under the square root is negative, there are no real solutions for this equation. Trigonometric functions can only output real values, so the equation has no real solutions for 't' that satisfy the given equation.