Question

5sin^2x-3sinxcosx=2cos^2x

Answer 1

Ответ:

Решение:

5 sin^2 x - 3 sin x cos x - 2 cos^2 x = 0.

делим на cos²x

5tg²x-3tgx-2=0

Пусть tgx=t

5t²-3t-2=0

t1=(3+7)/10=1

t2=-0.4

a) tgx=1

x1=π4+πn

б) tgx=-0.4

x2=-arctg0.4+πn

Объяснение:

Answer 2

To solve the trigonometric equation 5sin^2(x) - 3sin(x)cos(x) = 2cos^2(x), we can use trigonometric identities to simplify and eventually solve for x.

First, let's use the Pythagorean identity for sine and cosine:

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

Now, we can manipulate the equation to contain either sine or cosine terms only:

5(1 - cos^2(x)) - 3sin(x)cos(x) = 2cos^2(x)

Distribute the 5 on the left side:

5 - 5cos^2(x) - 3sin(x)cos(x) = 2cos^2(x)

Move all terms to one side of the equation:

5 - 5cos^2(x) - 3sin(x)cos(x) - 2cos^2(x) = 0

Combine like terms:

3 - 7cos^2(x) - 3sin(x)cos(x) = 0

Now, we can use the identity sin(2x) = 2sin(x)cos(x) to rewrite the equation:

3 - 7cos^2(x) - 3sin(2x) = 0

Now, we have a quadratic equation in terms of cos(x). Let's make a substitution:

Let y = cos(x).

The equation becomes:

3 - 7y^2 - 3sin(2x) = 0

Now, we can work with this equation in terms of y. We'll need to express sin(2x) in terms of y using the double-angle identity for sine:

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

Substitute this into the equation:

3 - 7y^2 - 3(2sin(x)y) = 0

3 - 7y^2 - 6y*sin(x) = 0

Now, you have a quadratic equation in y:

-7y^2 - 6y*sin(x) + 3 = 0

You can solve this quadratic equation for y. Once you find the values of y, you can use the inverse cosine function (arccos) to find the corresponding values of x.

Solving quadratic equations can be done using the quadratic formula:

y = [-b ± √(b² - 4ac)] / (2a)

In this case, a = -7, b = -6sin(x), and c = 3.

Solve for y, and then use arccos to find the values of x corresponding to the solutions for y. Keep in mind that this equation may have multiple solutions, and you should consider the domain of the trigonometric functions when determining the valid solutions for x.