sqrt (70)cosx+sqrt (70)sinx-sqrt (35)=0

To solve the equation $\sqrt{70}\cos x + \sqrt{70}\sin x - \sqrt{35} = 0$, we can use the trigonometric identity $\cos x + \sin x = \sqrt{2}\sin(x + \frac{\pi}{4})$.<br /><br />First, let's rewrite the equation as:<br />$\sqrt{70}(\cos x + \sin x) - \sqrt{35} = 0$<br /><br />Now, we can factor out $\sqrt{70}$ from the first term:<br />$\sqrt{70}(\cos x + \sin x) = \sqrt{35}$<br /><br />Divide both sides by $\sqrt{70}$:<br />$\cos x + \sin x = \frac{\sqrt{35}}{\sqrt{70}}$<br /><br />Simplify the right side:<br />$\cos x + \sin x = \frac{1}{\sqrt{2}}$<br /><br />Now, we can use the trigonometric identity $\cos x + \sin x = \sqrt{2}\sin(x + \frac{\pi}{4})$:<br />$\sqrt{2}\sin(x + \frac{\pi}{4}) = \frac{1}{\sqrt{2}}$<br /><br />Divide both sides by $\sqrt{2}$:<br />$\sin(x + \frac{\pi}{4}) = \frac{1}{2}$<br /><br />Now, we can find the values of $x$ that satisfy this equation. The general solution for $\sin(x) = \frac{1}{2}$ is:<br />$x = \frac{\pi}{6} + 2k\pi$ or $x = \frac{5\pi}{6} + 2k\pi$, where $k$ is an integer.<br /><br />So, the solutions for the given equation are:<br />$x = \frac{\pi}{6} + 2k\pi$ or $x = \frac{5\pi}{6} + 2k\pi$, where $k$ is an integer.