Question

16sinx-sin2x=1-cos2x

Answer 1

Будем использовать формулы:
sin2x=2sinx*cosx
cos2x =(cosx)^2 - (sinx)^2 
1 = (cosx)^2 + (sinx)^2

Решение:
16sinx-sin2x=1-cos2x
16sinx-2sinx*cosx=(cosx)^2 + (sinx)^2 - ((cosx)^2 - (sinx)^2)
16sinx-2sinx*cosx=(cosx)^2 + (sinx)^2 - (cosx)^2 + (sinx)^2
16sinx-2sinx*cosx=2 (sinx)^2 
8sinx-sinx*cosx -  (sinx)^2 =0
sinx*(8-cosx - sinx) =0
sinx = 0     или   8-cosx - sinx =0

sinx = 0
х = Пn, где n - целое число.

8-cosx - sinx =0
cosx + sinx =8 | $\frac{ \sqrt{2} }{2}$
$\frac{ \sqrt{2} }{2}$ cosx+ $\frac{ \sqrt{2} }{2}$sinx =  $4 \sqrt{2}$
cosx*sin(П/4) + sinx* cos(П/4) = $4 \sqrt{2}$
sin(П/4+x) =  $4 \sqrt{2}$
Данное уравнение НЕ имеет решений, Т.к. sin x не может быть больше 1

Ответ: х = Пn, где n - целое число.

Answer 2

To solve the trigonometric equation 16sinx - sin2x = 1 - cos2x, we'll need to use trigonometric identities to simplify the equation and find the value(s) of x that satisfy it.

Let's start by using the trigonometric identities:

1. sin2x = 2sinx*cosx
2. cos2x = 1 - 2sin^2x
3. 1 - cos2x = sin^2x

Now, let's substitute these identities into the original equation:

16sinx - 2sinx*cosx = sin^2x

To make it easier, we can bring everything to one side of the equation:

16sinx - 2sinx*cosx - sin^2x = 0

Now, the equation is in a quadratic form. We can factor it:

sinx(16 - 2cosx - sinx) = 0

Now, we have two possibilities to satisfy the equation:

1. sinx = 0
2. 16 - 2cosx - sinx = 0

Let's solve each possibility separately:

1. sinx = 0 If sinx = 0, this means x can be any integer multiple of π (pi). So, x = nπ, where n is an integer.

2. 16 - 2cosx - sinx = 0 Rearrange the equation: sinx = 16 - 2cosx

Now, use the Pythagorean identity sin^2x + cos^2x = 1: (16 - 2cosx)^2 + cos^2x = 1

Simplify and solve for cosx: 256 - 64cosx + 4cos^2x + cos^2x = 1 5cos^2x - 64cosx + 255 = 0

Now, solve this quadratic equation for cosx:

cosx = [64 ± √(64^2 - 45255)] / (2*5) cosx = [64 ± √(4096 - 5100)] / 10 cosx = [64 ± √(-1004)] / 10

Since the square root of a negative number is not a real number, there are no real solutions for cosx. Therefore, the second possibility does not provide real solutions for x.

In conclusion, the solutions to the original trigonometric equation are:

1. x = nπ, where n is an integer. (Infinite solutions)
2. There are no real solutions for the equation 16sinx - sin2x = 1 - cos2x when considering real values of x.