Question

Cos 3x =-1,6 2sin ^2x-3 sinxcosx =0

Answer 1

Cos 3x =-1,6   |cos3x|≤1   решения нет   x∈∅

2sin²x-3sinxcosx=0
sinx(2sinx-3cosx)=0
sinx=0   x=πn   n∈Z
2sinx=3cosx   tgx=1.5    x=arctg1.5+πn  n∈Z

Answer 2

To solve the trigonometric equations, let's consider each equation separately:

1. Cos(3x) = -1/6:

First, note that -1/6 is not within the range of the cosine function, which is between -1 and 1. Therefore, there are no real solutions to this equation.

2. 2sin^2(x) - 3sin(x)cos(x) = 0:

Factor out sin(x) from the equation:

sin(x) * (2sin(x) - 3cos(x)) = 0

Now, we have two possible cases:

Case 1: sin(x) = 0

This happens when x is an integer multiple of π (pi), i.e., x = nπ, where n is an integer.

Case 2: 2sin(x) - 3cos(x) = 0

To solve this equation, we can use the identity sin(2x) = 2sin(x)cos(x). Rearrange the equation:

2sin(x) - 3cos(x) = 2sin(x) - 3(2sin(x)cos(x)) = 0

Now, substitute sin(2x) into the equation:

2sin(x) - 6sin(2x) = 0

Now, use the double-angle identity for sine sin(2x) = 2sin(x)cos(x):

2sin(x) - 6(2sin(x)cos(x)) = 0

2sin(x) - 12sin(x)cos(x) = 0

Factor out sin(x) from the equation:

sin(x) * (2 - 12cos(x)) = 0

Now, we have two possible cases:

Case 2.1: sin(x) = 0

This was already considered in Case 1.

Case 2.2: 2 - 12cos(x) = 0

Solve for cos(x):

12cos(x) = 2

cos(x) = 2/12

cos(x) = 1/6

Now, to find the corresponding values of x, we can use the inverse cosine function:

x = ±cos^(-1)(1/6)

The solutions for x are:

x = cos^(-1)(1/6) and x = -cos^(-1)(1/6)

Please note that in Case 2, we found two possible solutions, but they need to be checked against the original equation to ensure they are valid. Make sure to verify them to confirm their validity.