Question

3sin^2 2x + 7cos2x - 3 = 0

Answer 1

3sin^2(2x)+7cos(2x)-3=0 {3=3sin^2(2x)+3cos^2(2x)}

3sin^2(2x)+7cos(2x)-3sin^2(2x)-3cos^2(2x)=0

7cos(2x)-3cos^2(2x)=0

cos(2x)(7-3cos(2x))=0

cos(2x)=0 или 7-3cos(2x)=0

2x=п/2+пn cos(2x)=7/3

x=п/4+пn/2 решений нет, т. к. 7/3>1

Пошаговое объяснение:

Answer 2

To solve the trigonometric equation 3sin^2(2x) + 7cos(2x) - 3 = 0, we'll first use the trigonometric identity:

sin^2(θ) + cos^2(θ) = 1

We can rewrite sin^2(2x) as 1 - cos^2(2x):

3(1 - cos^2(2x)) + 7cos(2x) - 3 = 0

Now, let's simplify the equation:

3 - 3cos^2(2x) + 7cos(2x) - 3 = 0

Next, combine like terms:

-3cos^2(2x) + 7cos(2x) = 0

Now, let's factor out the common term 'cos(2x)':

cos(2x)(-3cos(2x) + 7) = 0

Now, we have two cases to consider:

1. cos(2x) = 0
2. -3cos(2x) + 7 = 0

Case 1: cos(2x) = 0

To find the solutions for this case, set cos(2x) to zero and solve for x:

cos(2x) = 0

2x = π/2 + kπ or 2x = 3π/2 + kπ, where k is an integer

Divide by 2:

x = π/4 + kπ/2 or x = 3π/4 + kπ/2, where k is an integer

Case 2: -3cos(2x) + 7 = 0

Add 3cos(2x) to both sides:

7 = 3cos(2x)

Divide by 3:

cos(2x) = 7/3

The equation has no real solutions for this case since the cosine function ranges from -1 to 1, and 7/3 is outside this range.

Therefore, the solutions to the original trigonometric equation are:

x = π/4 + kπ/2, x = 3π/4 + kπ/2, where k is an integer.