Question

2cosx+cos2x=2sinx -как решить?

Answer 1

2cosx+cos2x=2sinx

2cosx+(2cos^2x-1)-2sinx=0

2cosx+2cos^2x-(sin^2x+cos^2x)-2sinx=0

2cosx+2cos^2x-sin^2x-cos^2x-2sinx=0
cos^2x+2cosx-sin^2x-2sinx=0

Произведём группировку

cos^2x-sin^2x+2cosx-2sinx=0

(cosx+sinx)(cosx-sinx)+2(cosx-sinx)=0

выносим общий множ. за скобки

(cosx-sinx)(cosx+sinx+2)=0

решаем по отдельности каждое ур-ие

 

1) cosx-sinx=0   /:cosx≠0

1-tgx=0

tgx=1

x=pi/4+pik, k ∈Z

 

2) cosx+sinx= - 2

√2(1/√2*cosx+1/√2*sinx)= - 2

sin(pi/4)cosx+cos(pi/4)*sinx= -2/√2

sin(pi/4+x)=-√2

-√2=1,41

нет реш. , т.к. x∈[-1;1]

 

ОТВЕТ:

pi/4+pik, k ∈Z

 

Answer 2

We can start by using the identity cos 2x = 1 - 2 sin^2 x to rewrite the equation as:

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

Simplifying and rearranging terms, we get:

2 cos x - 2 sin^2 x - 2 sin x + 1 = 0

Next, we can use the identity sin^2 x = 1 - cos^2 x to eliminate sin^2 x:

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

Simplifying again:

2 cos^2 x + 2 sin x - 1 = 0

Now we can use the quadratic formula to solve for cos x:

cos x = (-2 sin x ± sqrt(4 sin^2 x + 8))/4 cos x = (- sin x ± sqrt(sin^2 x + 2))/2

We can simplify further using the identity sin^2 x + cos^2 x = 1:

cos x = (- sin x ± sqrt(1 - cos^2 x + 2))/2 cos x = (- sin x ± sqrt(3 - cos^2 x))/2

To solve for cos x, we need to use some numerical methods, such as iteration or approximation, since there is no closed-form solution for this equation.