Question

144cos⁴ x-4sin⁴ x=9sin² 2x​

Answer 1

−4sin

2

x=4sin(

6

π

−x)sin(

6

π

+x),

1-4sin^{2}x = 4*\frac{1}{2}(cos(\frac{\pi}{6}-x-\frac{\pi}{6}-x)-cos(\frac{\pi}{6}-x+\frac{\pi}{6}+x)),1−4sin

2

x=4∗

2

1

(cos(

6

π

−x−

6

π

−x)−cos(

6

π

−x+

6

π

+x)),

1-4sin^{2}x=2cos2x-2cos\frac{\pi}{3},1−4sin

2

x=2cos2x−2cos

3

π

,

1-4sin^{2}x=2-4sin^{2}x-2*\frac{1}{2},1−4sin

2

x=2−4sin

2

x−2∗

2

1

,

1-4sin^{2}x-2+4sin^{2}x=-1,1−4sin

2

x−2+4sin

2

x=−1,

-1=-1.−1=−1.

ч. т. д.

Answer 2

Ответ:

если я правильно поняла то получиться х=0

Answer 3

To solve the trigonometric equation $144\cos^4 x - 4\sin^4 x = 9\sin^2 2x$, we can use trigonometric identities and algebraic manipulations. Let's break down the solution step by step.

Step 1: Use Double Angle Identity We'll start by using the double angle identity for sine: $\sin^2 2x = 2\sin x \cos x$. Substituting this into the equation gives: $144\cos^4 x - 4\sin^4 x = 9(2\sin x \cos x)$

Step 2: Use Pythagorean Identity Next, we can use the Pythagorean identity for sine and cosine: $\sin^2 x + \cos^2 x = 1$. Solving for $\cos^2 x$ gives us $\cos^2 x = 1 - \sin^2 x$. Substituting this into the equation gives: $144(1 - \sin^2 x)^2 - 4\sin^4 x = 18\sin x \cos x$

Step 3: Expand and Simplify Let's expand the squared term and simplify the equation: $144(1 - 2\sin^2 x + \sin^4 x) - 4\sin^4 x = 18\sin x \cos x$ $144 - 288\sin^2 x + 144\sin^4 x - 4\sin^4 x = 18\sin x \cos x$ $140\sin^4 x - 288\sin^2 x + 18\sin x \cos x - 144 = 0$

Step 4: Factor Let's factor out $\sin x$ from the third term: $140\sin^4 x - 288\sin^2 x + 18\sin x \cos x - 144 = \sin x (140\sin^3 x - 288\sin x + 18\cos x) - 144 = 0$

Step 5: Trig Identity Notice that the third term in the bracket can be expressed using the trig identity $\sin 3x = 3\sin x - 4\sin^3 x$: $140\sin^4 x - 288\sin^2 x + 18\sin x \cos x - 144 = \sin x (140\sin^3 x - 288\sin x + 18\cos x) - 144 = \sin x (140\sin^3 x - 144\sin x + 18\cos x)$

Step 6: Use Sine-Cosine Relationship Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can express $\cos x$ in terms of $\sin x$: $\cos x = \sqrt{1 - \sin^2 x}$

Substitute this expression into the equation: $\sin x (140\sin^3 x - 144\sin x + 18\cos x) = 0$ $\sin x (140\sin^3 x - 144\sin x + 18\sqrt{1 - \sin^2 x}) = 0$

Step 7: Solve for $\sin x$ We have two factors: $\sin x = 0$ and $140\sin^3 x - 144\sin x + 18\sqrt{1 - \sin^2 x} = 0$.

The first factor ($\sin x = 0$) gives us the solution $x = 0$ or $x = \pi$