Question

2cosx+sin2x=0 ............

Answer 1

2cosx+sin2x=0, заменим sin2x=2*sinx*cosx

2cosx+2sinxcosx=0, 2cosx вытащим за скобку и получим

2cosx(1+sinx)=0, приравниваем каждый многочлен на 0 и получим

2cosx=0 => x=π/2+πk, k€Z

1+sinx=0 => sinx=-1 => x=3π/2+2πn, n€Z

Answer 2

To solve the trigonometric equation 2cos(x) + sin(2x) = 0, you can use trigonometric identities and techniques. First, express sin(2x) in terms of sine and cosine using the double-angle identity for sine:

sin(2x) = 2sin(x)cos(x)

Now, substitute this expression into the original equation:

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

Now, factor out 2cos(x) from both terms:

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

Now, you have two factors to consider:

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

Let's solve each of these equations separately:

1. 2cos(x) = 0: Divide both sides by 2: cos(x) = 0

2. 1 + sin(x) = 0: Subtract 1 from both sides: sin(x) = -1

Now, let's find the solutions for each equation:

1. For cos(x) = 0, the solutions are x = π/2 + nπ, where n is an integer.

2. For sin(x) = -1, the solution is x = 3π/2 + 2nπ, where n is an integer.

So, the solutions to the original equation 2cos(x) + sin(2x) = 0 are:

x = π/2 + nπ (for the first equation) x = 3π/2 + 2nπ (for the second equation)

where n is an integer. These values of x satisfy the given equation.