Question

Cos3x+sinx sin2x=2cos^3x+2tgx

Answer 1

cos3x+sinx sin2x=2cos^3x+2tgx

cos3x роскладываеться на 4cos^3x-3cosx

будет 4cos^3x-3cosx+sinx*sin2x=2cos^3x+2tgx

sin2x=2sinxcosx

4cos^3x-2cos^3x -3cosx+2sin^2xcosx-2tgx=0

Выносим 2cosx с первого и третьего

2cos(cos^2x+sin^x)-3cosx-2tgx=0

cos^2x+sin^2x=1

2cosx-3cosx-2tgx=0

-cosx-2tgx=0

Роскладаем тангенс и домножаем обе части уровнения на 2cosx

-2sinx-2cos^2x=0

-2(sinx-cos^2x)=0

sinx-cos^2x=0

cos^2x=1-sin^2x

sinx+1-sin^2x=0

Пусть синикс=t

-t^2+t+1=0

t^2-t-1=0

D= 1-4*1*-1=5 =koren5^2

t1=1-koren5

        ---------

             2

 

t2=1+koren5

 -----------------

         2\

Answer 2

To simplify the given expression, let's work on each side separately:

Left side: cos(3x) + sin(x)sin(2x)

Using the trigonometric identity sin(2x) = 2sin(x)cos(x), we can substitute it into the expression:

cos(3x) + sin(x)(2sin(x)cos(x))

Expanding further:

cos(3x) + 2sin^2(x)cos(x)

Now, using the identity sin^2(x) = 1 - cos^2(x), we can rewrite the expression as:

cos(3x) + 2(1 - cos^2(x))cos(x)

Expanding again:

cos(3x) + 2cos(x) - 2cos^3(x)

Right side: 2cos^3(x) + 2tan(x)

Using the identity tan(x) = sin(x)/cos(x), we can substitute it into the expression:

2cos^3(x) + 2sin(x)/cos(x)

Combining the terms:

2cos^3(x) + 2sin(x)cos^(-1)(x)

Now, we can compare the simplified expressions on both sides:

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

By rearranging the terms:

cos(3x) + 2cos(x) = 4cos^3(x) + 2sin(x)cos^(-1)(x)

Unfortunately, I'm unable to solve this equation for a specific value of x without additional information or constraints. If you provide more details or specify any requirements, I can assist you further.