Question

Sqrt(2)*sin(2x+pi/4)-sqrt(2)sinx=sin2x+1 на отрезке[3pi/2;3pi]

Answer 1

Решение дано на фото.

Answer 2

To solve the trigonometric equation $\sqrt{2}\sin(2x+\frac{\pi}{4})-\sqrt{2}\sin(x) = \sin(2x)+1$ on the interval $[ \frac{3\pi}{2}, 3\pi]$, we'll first simplify the equation and then find the values of $x$ that satisfy it.

Step 1: Simplification Using trigonometric identities, we can simplify the equation:

$\sqrt{2}\sin(2x+\frac{\pi}{4}) - \sqrt{2}\sin(x) = \sin(2x) + 1$

Recall the angle sum formula for sine:

$\sin(A + B) = \sin A \cos B + \cos A \sin B$

Now let's apply this formula to the first term, where $A = 2x$ and $B = \frac{\pi}{4}$:

$\sin(2x + \frac{\pi}{4}) = \sin(2x)\cos(\frac{\pi}{4}) + \cos(2x)\sin(\frac{\pi}{4})$

Since $\cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, we have:

$\sin(2x + \frac{\pi}{4}) = \frac{1}{\sqrt{2}}\sin(2x) + \frac{1}{\sqrt{2}}\cos(2x)$

Now, substitute this back into the original equation:

$\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin(2x) + \frac{1}{\sqrt{2}}\cos(2x)\right) - \sqrt{2}\sin(x) = \sin(2x) + 1$

Cancel out the $\sqrt{2}$ terms:

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

Step 2: Rearrange the equation To simplify further, subtract $\sin(2x)$ from both sides:

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

Step 3: Use double-angle formula Now, recall the double-angle formulas for sine and cosine:

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

Substitute these into the equation:

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

Step 4: Simplify and solve Rearrange the equation:

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