Question

Sin^2x+15cosx+53=0 срочно

Answer 1

Дополним, используя основное тригонометрическое тождество:
1 - cos²x + 15cosx + 53 = 0
-cos²x + 15cosx + 54 = 0
cos²x - 15cosx - 54 = 0
Пусть t = cosx, t€[-1; 1].
t² - 15t - 54 = 0
t1 + t2 = 15
t1•t2 = -54

t1 = -18 - не входит в промежуток
t2 = 3 - не входит в промежуток.
Ответ: нет корней.

Answer 2

I'm sorry, but the equation you provided cannot be solved as it is.

The reason is that there is both a sine term and a cosine term in the equation, which makes it difficult to manipulate algebraically. Additionally, there is no information about what variable you are solving for (e.g. x, y, z).

If you meant to write the equation as:

sin^2(x) + 15cos(x) + 53 = 0

Then we can use some trigonometric identities to simplify it. Recall that sin^2(x) + cos^2(x) = 1, so we can substitute cos^2(x) = 1 - sin^2(x) into the equation:

sin^2(x) + 15cos(x) + 53 = 0 sin^2(x) + 15(1 - sin^2(x)) + 53 = 0 -14sin^2(x) + 15 = 0

Now we can solve for sin(x):

-14sin^2(x) + 15 = 0 sin^2(x) = 15/14 sin(x) = ±√(15/14)

Since sine is positive in the first and second quadrants, the solutions are:

x = arcsin(√(15/14)) ≈ 0.961 radians x = π - arcsin(√(15/14)) ≈ 2.181 radians

Note that there is no real solution for cosine since the cosine term does not contain a square.