Math, asked by abhishekraj58, 11 months ago

cos2x - 5sinx - 3 = 0​

Answers

Answered by arkaroy45
1

Step-by-step explanation:

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

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

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

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

2(sin(x))^2 + 5sin(x) + 2 = 0

'

(2sin(x)) + 1)(sin(x) + 2) = 0

solution occurs when 2sin(x) + 1 = 0... sin(x) + 2 can never equal 0 since that would require sin(x) = -2 (that is not possible)...

2sin(x) + 1 = 0

2sin(x) = - 1

sin(x) = -1/2

between 0 and 360 degrees, this shall be satisfied at 210 degrees and 330 degrees... more generally, the solution could be expressed in degrees as (210 + 360(n-1)) degrees and (330 + 360(n-1)) degrees where n runs from 1,2,3.........

Answered by shadowsabers03
8

     cos(2x) - 5sin (x) - 3 = 0​  

⇒   1 - 2 sin² (x) - 5 sin (x) - 3 = 0  

⇒  - 2 sin² (x) - 5 sin (x) - 2 = 0  

⇒  - 2 sin² (x) - 5 sin (x) - 2 = 0  

⇒  2 sin² (x) + 5 sin (x) + 2 = 0  

⇒  2 sin² (x) + sin (x) + 4 sin (x) + 2 = 0  

⇒  sin (x) [2 sin (x) + 1] + 2 [2 sin (x) + 1] = 0  

⇒  [2 sin (x) + 1] [sin (x) + 2] = 0  

⇒  sin (x) = -1/2          ;          sin (x) = -2  

But since sin (x) ∈ [-1, 1], sin (x) ≠ 2, because -2 ∉ [-1, 1].  

So, sin (x) = -1/2.  

Since sin (x) is negative, x lies in third or fourth quadrant.    

We know sin (π/6) = 1/2. Consider this one.  

We will get RHS as -1/2 if we add π or 3π/2 to π/6. → (1)  

We will get RHS as -1/2 if we subtract π/2 or π from π/6 too. → (2)  

From (1) and (2), we can see a common thing.  

sin (π/6 ± π) = -1/2  

On adding or subtracting 2π continuously to/from π/6 ± π, the RHS won't change.  

sin (π/6 ± π ± 2π ± 2π ± 2π ±...) = -1/2  

We simply write this as,  

sin (π/6 + (2n - 1)π) = -1/2  ∀n ∈ Z.  

From this we get,  

x = π/6 + (2n - 1)π = (2n - 5/6)π,  ∀n ∈ Z.    

From (1) and (2), we can see another common thing.        

sin (π/6 + π ± π/2) = -1/2  

⇒  sin (7π/6 ± π/2) = -1/2  

As we did earlier, on adding or subtracting 2π continuously to/from 7π/6 ± π/2, the RHS won't change.  

sin (7π/6 ± π/2 ± 2π ± 2π ± 2π ±...) = -1/2  

We simply write this as,  

sin (7π/6 + (4n + 1)π/2) = -1/2, ∀n ∈ Z.  

From this, we get,  

x = 7π/6 + (4n + 1)π/2 = (6n + 5)π/3, ∀n ∈ Z.  

So, two possible values for x are found.  

1.  x = (2n - 5/6)π  

2.  x = (6n + 5)π/3

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