Question

(sinx-sin3x)/1-cosx=0 [-P/2;P/2]

Answer 1

Сначала я применю формулу тройного синуса, потом подготовлю к методу интервалов. И в ответе укажу значения только для данного промежутка.

$\frac{sinx-sin3x}{1-cosx}=0\\ \frac{2sinx(2sin^2x-1)}{cox-1} =0\\\frac{sinx(sinx+\frac{\sqrt{2} }{2} )(sinx-\frac{\sqrt{2} }{2} )}{cosx-1}=0\\\left \{ {{x\neq 2pi*n} \atop {\left[\begin{array}{ccc}x=pi*n\\x=\frac{pi}{4}+\frac{pi}{2}*n \\\end{array}}} \right.$

Ответ: x=±pi/4

Answer 2

To solve the equation $\frac{{\sin{x} - \sin{3x}}}{{1 - \cos{x}}} = 0$ over the interval $[- \frac{\pi}{2}, \frac{\pi}{2}]$, we need to find the values of $x$ that make the expression equal to zero.

Step 1: Simplify the expression To simplify the expression, we can use the identity $\sin{3x} = 3\sin{x} - 4\sin^3{x}$:

$\frac{{\sin{x} - (3\sin{x} - 4\sin^3{x})}}{{1 - \cos{x}}} = 0$

Step 2: Further simplify the expression Combine like terms:

$\frac{{-2\sin{x} + 4\sin^3{x}}}{{1 - \cos{x}}} = 0$

Step 3: Factor out $2\sin{x}$:

$\frac{{2\sin{x}(2\sin^2{x} - 1)}}{{1 - \cos{x}}} = 0$

Step 4: Use trigonometric identities Now, we can use the trigonometric identity $\sin^2{x} + \cos^2{x} = 1$ to replace $1 - \cos{x}$ in the denominator:

$\frac{{2\sin{x}(2\sin^2{x} - 1)}}{{\sin^2{x}}} = 0$

Step 5: Eliminate the fraction To eliminate the fraction, we can multiply both sides by $\sin^2{x}$ (but we need to be careful and ensure that $\sin^2{x}$ is not zero):

$\begin{cases} 2\sin{x}(2\sin^2{x} - 1) = 0, & \text{if } \sin^2{x} \neq 0 \\ 0 = 0, & \text{if } \sin^2{x} = 0 \end{cases}$

Step 6: Solve for $\sin{x}$ Now, we have two cases:

Case 1: $\sin^2{x} \neq 0$ In this case, we can divide both sides by $\sin{x}$:

$2(2\sin^2{x} - 1) = 0$

Expand and rearrange:

$4\sin^2{x} - 2 = 0$

Solve for $\sin^2{x}$:

$4\sin^2{x} = 2$

$\sin^2{x} = \frac{1}{2}$

Since we are looking for solutions in the interval $[- \frac{\pi}{2}, \frac{\pi}{2}]$, the possible solutions for $x$ are $\frac{\pi}{4}$ and $-\frac{\pi}{4}$ because $\sin{\frac{\pi}{4}} = \sin{-\frac{\pi}{4}} = \frac{\sqrt{2}}{2}$.

Case 2: $\sin^2{x} = 0$ If $\sin^2{x} = 0$, then $\sin{x} = 0$. The only solution within the given interval is $x = 0$.

Step 7: Final solutions The solutions within the given interval are $x = 0$, $x = \frac{\pi}{4}$, and $x = -\frac{\pi}{4}$