Question

.sin2x - 3 sinx cosx + 2cosx= 0.​

Answer 1

Ответ:

x=п/2+пn

Объяснение:

sin2x - 3 sinx cosx + 2cosx= 0.

2sinxcosx - 3sinxcosx+2cosx=0

2cosx-sinxcosx=0

cosx(2-sinx)=0

cosx=0 sinx=2 - не удовлетв.

x=п/2+пn т.к sinx<=1

Answer 2

To solve the trigonometric equation .sin(2x) - 3sin(x)cos(x) + 2cos(x) = 0, we can use some trigonometric identities and manipulation techniques. Let's go step by step:

Step 1: Use the double-angle identity for sine: sin(2x) = 2sin(x)cos(x).

Now, the equation becomes: 2sin(x)cos(x) - 3sin(x)cos(x) + 2cos(x) = 0.

Step 2: Factor out common terms: We can factor out sin(x)cos(x) from the equation.

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

Step 3: Simplify the equation:

-sin(x)cos(x) + 2cos(x) = 0.

Step 4: Factor out the common term, cos(x):

cos(x) * (-sin(x) + 2) = 0.

Now, we have two possible cases to consider:

Case 1: cos(x) = 0.

Case 2: -sin(x) + 2 = 0.

Let's solve each case:

Case 1: cos(x) = 0.

If cos(x) = 0, then x can take the values of (π/2 + πk) or (3π/2 + πk), where k is an integer.

Case 2: -sin(x) + 2 = 0.

Adding sin(x) to both sides:

sin(x) = 2.

However, sin(x) can never exceed 1 in magnitude, so there is no real solution for this case.

Therefore, the solutions to the original equation are:

1. x = π/2 + πk, where k is an integer.
2. x = 3π/2 + πk, where k is an integer.