Question

Sin2x*cosx-sinx+cos2x=0

Answer 1

Sin2x*cosx-sinx+cos2x=0
2cosx*sinx*cosx-sinx+cos2x=0
2cos²x*sinx-sinx+cos2x=0
2*(1+cos2x)/2*sinx-sinx+cos2x=0
(cos2x+1)*sinx-sinx+cos2x=0
sinx+cos2x*sinx-sin+cos2x=0
cos2x*sinx+cos2x=0
cos2x(sinx+1)=0
cos2x=0 или sinx+1=0
cos2x=0, при 2x=90°, x=45
sinx+1=0
sinx=-1
при x=270

Answer 2

To solve the trigonometric equation $\sin(2x)\cdot\cos(x) - \sin(x) + \cos(2x) = 0$, we'll need to use some trigonometric identities and algebraic manipulations. Here's how you can do it step by step:

Step 1: Use trigonometric identities to rewrite $\sin(2x)$ and $\cos(2x)$. $\sin(2x) = 2\sin(x)\cos(x)$ $\cos(2x) = \cos^2(x) - \sin^2(x)$

Step 2: Substitute the above expressions into the original equation: $2\sin(x)\cos(x)\cdot\cos(x) - \sin(x) + (\cos^2(x) - \sin^2(x)) = 0$

Step 3: Simplify the equation: $2\sin(x)\cos^2(x) - \sin(x) + \cos^2(x) - \sin^2(x) = 0$

Step 4: Rearrange the terms and factor if possible: $\cos^2(x)(2\sin(x) + 1) - \sin^2(x) - \sin(x) = 0$

Step 5: Apply a trigonometric identity: $\sin^2(x) + \cos^2(x) = 1$ $(2\sin(x) + 1)\cos^2(x) - (1 - \cos^2(x))\sin(x) = 0$

Step 6: Further simplify: $2\sin(x)\cos^2(x) + \cos^2(x) - \sin(x) = 0$

Step 7: Factor out $\cos^2(x)$: $\cos^2(x)(2\sin(x) + 1) - \sin(x) = 0$

Step 8: Set each factor equal to zero and solve for $x$:

1. $\cos^2(x) = 0$ If $\cos^2(x) = 0$, then $\cos(x) = 0$. The solutions for this are $x = \frac{\pi}{2} + n\pi$ and $x = \frac{3\pi}{2} + n\pi$ where $n$ is an integer.

2. $2\sin(x) + 1 = 0$ If $2\sin(x) + 1 = 0$, then $\sin(x) = -\frac{1}{2}$. The solutions for this are $x = \frac{7\pi}{6} + 2n\pi$ and $x = \frac{11\pi}{6} + 2n\pi$ where $n$ is an integer.

Therefore, the complete set of solutions for the equation $\sin(2x)\cdot\cos(x) - \sin(x) + \cos(2x) = 0$ is: $x = \frac{\pi}{2} + n\pi, \quad x = \frac{3\pi}{2} + n\pi, \quad x = \frac{7\pi}{6} + 2n\pi, \quad x = \frac{11\pi}{6} + 2n\pi$ where $n$ is an integer.