Question

Sin 2x =2 sin x -cos x +1

Answer 1

Ответ:

Решение:  

1) Sin (x-1) = cos (x+2)  

Sin (x-1) - cos (x+2)=0  

Sin (x-1) - sin(π/2- (x+2))=0  

2sin(π/4-3/2)*cos(x+1/2-π/4)=0  

cos(x+1/2-π/4)=0  

x+1/2-π/4=π/2+πn  

x=π/2-1/2+π/4+πn  

x=3π/2-1/2+πn  

2) Sin x Sin 5x = Cos 4x;  

1/2(cos(-4x)-cos6x)-cos4x=0  

cos4x-cos6x-2cos4x=0  

cos6x+cos4x=0  

2cos5x*cosx=0  

a) cos5x=0  

5x=π/2+πn  

x1=π/10+πn/5  

б) cosx=0  

x2=π/2+πn

Пошаговое объяснение:

Answer 2

To solve the trigonometric equation sin(2x) = 2sin(x) - cos(x) + 1, we'll first try to rewrite the equation using trigonometric identities. Then we'll solve for x. Here's the step-by-step solution:

Step 1: Use the double-angle identity for sine (sin(2x) = 2sin(x)cos(x)) and the Pythagorean identity (sin^2(x) + cos^2(x) = 1) to simplify the equation:

2sin(x)cos(x) = 2sin(x) - cos(x) + 1

Step 2: Move all the terms to one side of the equation:

2sin(x)cos(x) - 2sin(x) + cos(x) - 1 = 0

Step 3: Factor out common terms:

sin(x)(2cos(x) - 2) + cos(x) - 1 = 0

Step 4: Factor out 2 from the first term inside the parenthesis:

sin(x)(2(cos(x) - 1)) + cos(x) - 1 = 0

Step 5: Now, we have a common factor of (2cos(x) - 2) between the first two terms, so let's factor that out:

2(cos(x) - 1)(sin(x) + 1) = 0

Step 6: Now, set each factor equal to zero and solve for x:

1. 2 = 0: This is not possible, as 2 is not equal to zero.

2. cos(x) - 1 = 0: cos(x) = 1 x = 0° + 360°n (n is an integer)

3. sin(x) + 1 = 0: sin(x) = -1 x = 270° + 360°n (n is an integer)

So, the solutions for the equation sin(2x) = 2sin(x) - cos(x) + 1 are: x = 0° + 360°n (n is an integer) x = 270° + 360°n (n is an integer)