Question

Решить уравнение. 8cos^2 x+3sin2x-7=0 (В ЛС ПИШИТЕ У МЕНЯ ССЫЛКА НА ЕЩЕ ОДНО МОЕ ЛЕГЕНЬКОЕ ЗАДАНИЕ)

Answer 1

8cos^2x+6sinxcosx-7cos^2x-7sin^2x=0
1+6tgx-7tg^2x=0
-7t^2+6t+1=0
7t^2-6t-1=0
D=64
Дальше просто. 

Answer 2

To solve the equation 8cos^2(x) + 3sin(2x) - 7 = 0, we can use trigonometric identities and algebraic manipulation. Let's break it down step by step:

Step 1: Simplify the equation using trigonometric identities

We can use the double-angle identity for sine: sin(2x) = 2sin(x)cos(x). Substituting this into the equation, we get:

8cos^2(x) + 3(2sin(x)cos(x)) - 7 = 0

Simplifying further:

8cos^2(x) + 6sin(x)cos(x) - 7 = 0

Step 2: Use the Pythagorean identity to eliminate cos^2(x)

The Pythagorean identity states that sin^2(x) + cos^2(x) = 1. Rearranging this equation, we get cos^2(x) = 1 - sin^2(x). Substituting this into our equation, we have:

8(1 - sin^2(x)) + 6sin(x)cos(x) - 7 = 0

Simplifying further:

8 - 8sin^2(x) + 6sin(x)cos(x) - 7 = 0

Step 3: Rearrange the equation

Let's rearrange the equation to group the terms:

-8sin^2(x) + 6sin(x)cos(x) + 1 = 0

Step 4