Question

Sin6x + cos6x=1- 3/4 sin2 2x​

Answer 1

[то что синим начертано] - y=sin(6x)+cos(6x)

[то что зелёным начертано] - y=1-3/4×sin(2)×2x

Объяснение:

Область определения функции : xєR

Пересечение с осью ординат : (0;1)

Корень : (2/3sin(2),0)

Answer 2

To simplify the given trigonometric expression, we can use trigonometric identities to rewrite the terms on both sides of the equation and then solve for x.

Given equation: sin(6x) + cos(6x) = 1 - (3/4) * sin^2(2x)

First, let's focus on the right-hand side of the equation and simplify it: 1 - (3/4) * sin^2(2x) = 1 - (3/4) * (1 - cos^2(2x)) [Using the identity: sin^2(x) = 1 - cos^2(x)] = 1 - (3/4) + (3/4) * cos^2(2x) = 1 - 3/4 + (3/4) * cos^2(2x)

Now, the equation becomes: sin(6x) + cos(6x) = 1 - 3/4 + (3/4) * cos^2(2x)

Let's simplify further: sin(6x) + cos(6x) = 1/4 + (3/4) * cos^2(2x)

Now, we can use the trigonometric identity: cos^2(x) = 1 - sin^2(x) cos^2(2x) = 1 - sin^2(2x)

Substitute this into the equation: sin(6x) + cos(6x) = 1/4 + (3/4) * (1 - sin^2(2x)) sin(6x) + cos(6x) = 1/4 + 3/4 - (3/4) * sin^2(2x) sin(6x) + cos(6x) = 1 - (3/4) * sin^2(2x)

As you can see, the right-hand side of the given equation matches the simplified expression from earlier. Therefore, the original equation is true for all values of x.

In conclusion, the equation sin(6x) + cos(6x) = 1 - (3/4) * sin^2(2x) holds true for all values of x.