Question

Sin 3x =-1/2 -2cos x=1

Answer 1

Вот ответ, но не очень уверена

Answer 2

To solve the equation sin(3x) = -1/2 - 2cos(x) = 1, we'll break it down into two separate equations and solve them individually. Let's start with sin(3x) = -1/2:

1. Solve sin(3x) = -1/2: To find the values of x that satisfy this equation, we can use the inverse sine function (arcsin or sin^(-1)).

arcsin(-1/2) = -π/6 + 2πk or 7π/6 + 2πk, where k is an integer.

So, we have two cases: Case 1: 3x = -π/6 + 2πk Case 2: 3x = 7π/6 + 2πk

2. Solve -2cos(x) = 1: To solve this equation, we'll isolate cos(x) and find its value.

-2cos(x) = 1 cos(x) = -1/2

Using the inverse cosine function (arccos or cos^(-1)), we find the solutions:

arccos(-1/2) = 2π/3 + 2πk or 4π/3 + 2πk, where k is an integer.

Now, we'll combine the solutions from both equations and find the common values for x:

For Case 1: 3x = -π/6 + 2πk For Case 2: 3x = 7π/6 + 2πk

For cos(x) = -1/2: x = 2π/3 + 2πk or x = 4π/3 + 2πk

Now, we need to find the values of x that satisfy both conditions.

For x = 2π/3 + 2πk: 3(2π/3 + 2πk) = -π/6 + 2πk 2π + 6πk = -π/6 + 2πk

Simplifying, we get: 4π + 12πk = -π + 12πk 4π = -π

This equation has no solutions.

For x = 4π/3 + 2πk: 3(4π/3 + 2πk) = -π/6 + 2πk 4π + 12πk = -π/6 + 2πk

Simplifying, we get: 8π + 24πk = -π + 12πk 8π = -π

This equation has no solutions.

In conclusion, there are no values of x that satisfy the equation sin(3x) = -1/2 - 2cos(x) = 1.