Question

1 +3 sin ^2x = 2sin 2x

Answer 1

Решение: 
cos²x+sin²x+3sin²x-4sinxcosx=0 
cos²x+4sin²x-4sinxcosx=0 (делим обе части на cos²x не=0) 
4tg²x-4tgx+1=0 
tgx=y,4y²-4y+1=0 
(2y-1)²=0 
2y-1=0 
y=1/2 
tgx=1/2 
x=arctg(1/2)+pin,n€Z.

Answer 2

Let's solve the trigonometric equation 1 + 3sin^2(x) = 2sin(2x) step by step:

Step 1: Simplify the equation using trigonometric identities. Recall the double angle identity for sine: sin(2x) = 2sin(x)cos(x).

The equation becomes: 1 + 3sin^2(x) = 2(2sin(x)cos(x))

Step 2: Expand the right side of the equation: 1 + 3sin^2(x) = 4sin(x)cos(x)

Step 3: We can use the Pythagorean identity: sin^2(x) + cos^2(x) = 1, to express cos(x) in terms of sin(x): cos^2(x) = 1 - sin^2(x) cos(x) = √(1 - sin^2(x))

Step 4: Substitute the expression for cos(x) into the equation: 1 + 3sin^2(x) = 4sin(x)√(1 - sin^2(x))

Step 5: To simplify further, let's square both sides to eliminate the square root: (1 + 3sin^2(x))^2 = (4sin(x)√(1 - sin^2(x)))^2

Step 6: Expand both sides: 1 + 6sin^2(x) + 9sin^4(x) = 16sin^2(x)(1 - sin^2(x))

Step 7: Bring all terms to one side of the equation: 9sin^4(x) + 6sin^2(x) + 1 - 16sin^2(x) + 16sin^2(x) = 0

Step 8: Simplify: 9sin^4(x) - 6sin^2(x) + 1 = 0

Step 9: Factor the quadratic equation. Let u = sin^2(x): 9u^2 - 6u + 1 = 0

Step 10: Solve for u using the quadratic formula (or factoring): u = [6 ± √(6^2 - 4 * 9 * 1)] / (2 * 9) u = [6 ± √(36 - 36)] / 18 u = [6 ± √(0)] / 18 u = [6 ± 0] / 18

There are two solutions for u: u = 6 / 18 = 1/3 u = 6 / 18 = 1/3

Step 11: Solve for sin(x): Since u = sin^2(x), we take the square root of each solution for u to find the solutions for sin(x):

1. sin(x) = √(1/3) = √3/3
2. sin(x) = -√(1/3) = -√3/3

So the solutions for the original equation 1 + 3sin^2(x) = 2sin(2x) are:

1. sin(x) = √3/3
2. sin(x) = -√3/3