Question

Найдите сумму корней уравнения из промежутка (-пи; пи), учитывая, что это уравнение следующее:a) 3(1-sinx)=1+cos2xb) 3sin2x+ 8cos^2 x=1За подробное решение- лучший ответ!

Answer 1

a)

cos2x = cos^2(x)-sin^2(x) = (1-sin^2(x)) - sin^2(x) = 1 - 2sin^2(x)

3(1-sinx) - 1- cos2x =0  ==> 3 - 3sinx - 1- (1 - 2sin^2(x) = 0  ==> 3 - 3sinx - 2 + 2sin^2(x) = 0

2sin^2(x) - 3sinx +1 = 0

обозначим y=sinx  тогда получим квадратное уравнение

2y^2 - 3y +1 = 0  корни которого y1=1 и y2=1/2  

y1=1  ==> sinx =1  ==>  x1=pi/2

y2=1/2  ==> sinx =1/2  ==> 1)  x2=pi/6 

                                                2)  x3=pi-pi/6 =5pi/6

 

x1+x2+x3=pi/2+pi/6+5pi/6=3*pi/2

b)  

sin2x = 2sinx*cosx       1=cos^2 x + sin^2 x

3sin2x+ 8cos^2 x - 1 =0  ==> 3(2sinx*cosx) + 8cos^2 x  -(cos^2 x + sin^2 x)=0  ==> 

6sinx*cosx + 7cos^2 x  - sin^2 x =0   раздедим на  (-cos^2 x) получим

-6tgx - 7 + tg^2 x =0  ==>  tg^2 x  - 6tgx - 7=0

обозначим y=tgx  тогда получим квадратное уравнение

y^2  - 6y - 7=0

D= 36-4*1*(-7) =36+28 =64 = 8^2

y1=(6+8)/(2*1)=14/2 =7

y2=(6-8)/(2*1)=-2/2 =-1

 

y1=7   tgx=7  ==>  x1 = arctg(7)

y2=-1   tgx=-1  ==>  1) x2 = -pi/4

                                   2) x3 = 3*pi/4

 

x1+x2+x3=arctg(7)-pi/4+3*pi/4=pi/2 + arctg(7)

Answer 2

Solution for Equation a)

To find the sum of the roots of the equation 3(1-sin(x)) = 1 + cos(2x) over the interval (-π, π), we can follow these steps:

Step 1: Simplify the equation: - Distribute the 3 on the left side: 3 - 3sin(x) = 1 + cos(2x) - Rearrange the terms: 3sin(x) + cos(2x) = 2

Step 2: Use the double angle formula for cosine: - The double angle formula for cosine states that cos(2x) = 2cos^2(x) - 1 - Substitute this into the equation: 3sin(x) + 2cos^2(x) - 1 = 2

Step 3: Rearrange the equation: - Move all terms to one side: 2cos^2(x) + 3sin(x) - 3 = 0

Step 4: Solve the equation: - Unfortunately, there is no simple algebraic method to find the roots of this equation. We can use numerical methods or graphing to approximate the roots.

Solution for Equation b)

To find the sum of the roots of the equation 3sin(2x) + 8cos^2(x) = 1 over the interval (-π, π), we can follow these steps:

Step 1: Simplify the equation: - Rearrange the terms: 8cos^2(x) + 3sin(2x) = 1

Step 2: Use the double angle formula for sine: - The double angle formula for sine states that sin(2x) = 2sin(x)cos(x) - Substitute this into the equation: 8cos^2(x) + 6sin(x)cos(x) = 1

Step 3: Rearrange the equation: - Move all terms to one side: 8cos^2(x) + 6sin(x)cos(x) - 1 = 0

Step 4: Solve the equation: - Unfortunately, there is no simple algebraic method to find the roots of this equation. We can use numerical methods or graphing to approximate the roots.

In both cases, we need to use numerical methods or graphing to approximate the roots and find their sum over the given interval.