Question

Sin3x*cos4x+cos3x*sin4x=√3\2​

Answer 1

$\sin 3x\cos 4x+\cos 3x\sin4x=\dfrac{\sqrt{3}}{2}\\ \\ \sin(3x+4x)=\dfrac{\sqrt{3}}{2}\\ \\ \sin7x=\dfrac{\sqrt{3}}{2}\\ \\ 7x=(-1)^k\cdot \dfrac{\pi}{3}+\pi k,k \in \mathbb{Z}\\ \\ x=(-1)^k\cdot \dfrac{\pi}{21}+\dfrac{\pi k}{7},k \in \mathbb{Z}$

Answer 2

To solve the equation sin(3x)*cos(4x) + cos(3x)*sin(4x) = √3/2, we can use the trigonometric identity sin(A + B) = sin(A)*cos(B) + cos(A)*sin(B).

Applying the identity, we have:

sin(3x + 4x) = √3/2

Simplifying:

sin(7x) = √3/2

Now, we need to find the values of x that satisfy this equation.

The equation sin(7x) = √3/2 corresponds to the angle whose sine is √3/2, which is π/3. Since sine has a period of 2π, we can add any multiple of 2π to the angle to find all solutions.

Therefore, we have:

7x = π/3 + 2πn (where n is an integer)

Simplifying:

x = (π/3 + 2πn)/7

So, the solutions to the equation sin(3x)*cos(4x) + cos(3x)*sin(4x) = √3/2 are given by:

x = (π/3 + 2πn)/7, where n is an integer.