Question

2 + sinxcosx = 2sinx + cosx

Answer 1

$2+\sin x\cos x=2\sin x+\cos x \\ 2-2\sin x+\sin x\cos x-\cos x=0 \\ 2(1-\sin x)-\cos x(1-\sin x)=0 \\ (1-\sin x)(2-\cos x)=0 \\ \left[\begin{array}{ccc}\sin x=1\\ \cos x=2\end{array}\right\to \left[\begin{array}{ccc}x= \frac{\pi}{2}+2 \pi k,k \in Z \\ \O \end{array}\right$

Answer 2

To solve the equation 2 + sin(x)cos(x) = 2sin(x) + cos(x), you can rearrange the terms and apply some trigonometric identities. Here's how to do it:

1. Move all the terms to one side of the equation to set it equal to zero: 2 + sin(x)cos(x) - 2sin(x) - cos(x) = 0

2. Now, let's use some trigonometric identities:

   • Rewrite sin(x)cos(x) as (1/2)sin(2x).
   • Rewrite 2sin(x) as sin(x)cos(x) + sin(x)cos(x) = 2sin(x)cos(x).

The equation now becomes: 2 + (1/2)sin(2x) - 2sin(x) - cos(x) = 0

3. Combine like terms: 2sin(x)cos(x) + (1/2)sin(2x) - 2sin(x) - cos(x) - 2 = 0

4. Now, let's factor out common terms: 2sin(x)cos(x) - 2sin(x) + (1/2)sin(2x) - cos(x) - 2 = 0

5. Factor out sin(x) from the first two terms and cos(x) from the last two terms: 2sin(x)(cos(x) - 1) + (1/2)sin(2x) - (cos(x) + 2) = 0

6. Notice that you can apply the double angle identity for sin(2x) = 2sin(x)cos(x): 2sin(x)(cos(x) - 1) + (1/2)(2sin(x)cos(x)) - (cos(x) + 2) = 0

7. Continue simplifying: 2sin(x)(cos(x) - 1) + sin(x)cos(x) - (cos(x) + 2) = 0

8. Combine like terms: 3sin(x)cos(x) - 2cos(x) - 2 = 0

9. Now, factor out common terms: cos(x)(3sin(x) - 2) - 2 = 0

10. Set each factor equal to zero: cos(x) = 0 (1) 3sin(x) - 2 = 0 (2)

Now, you have two equations to solve. Equation (1) yields cos(x) = 0, which means x = π/2 + kπ, where k is an integer.

For Equation (2), solve for sin(x): 3sin(x) = 2 sin(x) = 2/3

Now, find the values of x for sin(x) = 2/3. You can use the inverse sine function:

x = arcsin(2/3)

So, you have two sets of solutions:

1. x = π/2 + kπ (where k is an integer)
2. x = arcsin(2/3)

These are the solutions to the equation 2 + sin(x)cos(x) = 2sin(x) + cos(x).