Question

Sin 5x + sin x + 2 cos^2 x = 1 ____________________________________

Answer 1

2sin(3x)*cos (2x)+2(1-cos(2x))/2=1. 2sin(3x)*cos(2x)+1-cos(2x)=1 . cos(2x)=1.  cos(2x)(2sin/3x)-1)=0.   2sin(3x)-1=0.  2x=n/2+ni.   x=n/4+n/2.   sin(3x)=1/2.   3x=(-1)*n/6+nm.     x=(-1)*n/18+nm/3

Answer 2

To solve the trigonometric equation $\sin(5x) + \sin(x) + 2\cos^2(x) = 1$, we'll use trigonometric identities and algebraic manipulation.

First, let's rewrite $\cos^2(x)$ in terms of $\sin(x)$ using the identity $\sin^2(x) + \cos^2(x) = 1$:

$\cos^2(x) = 1 - \sin^2(x)$

Now, let's substitute this back into the original equation:

$\sin(5x) + \sin(x) + 2(1 - \sin^2(x)) = 1$

Simplify further:

$\sin(5x) + \sin(x) + 2 - 2\sin^2(x) = 1$

Rearrange terms:

$-2\sin^2(x) + \sin(5x) + \sin(x) + 1 = 0$

Since we have a quadratic term in $\sin(x)$, it might be helpful to rewrite everything in terms of $\sin(2x)$ using the double-angle identity:

$\sin(2x) = 2\sin(x)\cos(x)$

We can express $\cos(x)$ in terms of $\sin(x)$ using $\sin^2(x) + \cos^2(x) = 1$:

$\cos(x) = \sqrt{1 - \sin^2(x)}$

$\cos(x) = \pm \sqrt{1 - \sin^2(x)}$ (since $\cos(x)$ can be positive or negative depending on the quadrant)

Now, rewrite $\sin(5x)$ in terms of $\sin(2x)$ using the multiple-angle formula:

$\sin(5x) = 16\sin^5(x) - 20\sin^3(x) + 5\sin(x)$

Substitute $\sin(2x)$ and $\cos(x)$ back into the equation:

$-2\sin^2(x) + (16\sin^5(x) - 20\sin^3(x) + 5\sin(x)) + \sin(x) + 1 = 0$

Simplify further:

$16\sin^5(x) - 20\sin^3(x) - \sin^2(x) + 6\sin(x) + 1 = 0$

At this point, it's a quintic equation in terms of $\sin(x)$, which is generally not straightforward to solve analytically. You may need to resort to numerical methods or specific techniques for solving quintic equations if an exact solution is required. Keep in mind that solutions may be complex, and not all equations are guaranteed to have real solutions.