Solution-4

We will prove the following inequality: \(a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\). Raise both sides of the inequality to the power two and use the identity \(\sin ^{2} x+\cos ^{2} x=1\). We get the following inequalities:

\[\begin{gathered} (a \sin x+b \cos x)^{2} \leq\left(a^{2}+b^{2}\right)\left(\sin ^{2} x+\cos ^{2} x\right) \\ a^{2} \sin ^{2} x+2 a b \sin x \cos x+b \cos ^{2} x \leq a^{2} \sin ^{2} x+a^{2} \cos ^{2} x+b^{2} \sin ^{2} x+b^{2} \cos ^{2} x \\ 0 \leq a^{2} \cos ^{2} x-2 a b \sin x \cos x+b^{2} \sin ^{2} x \\ (a \cos x-b \sin x)^{2} \geq 0 \end{gathered}\]

The last inequality is true for any \(a, b\) and \(x\). Therefore the initial inequality \(a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\) is true as well. It follows that \(3 \sin x+4 \cos x \leq \sqrt{3^{2}+4^{2}}=5\), and therefore the original equation has no roots.

Atrisinājums-4

Pierādīsim, ka izpildās nevienādība \(a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\). Kāpināsim abas nevienādības puses kvadrātā un izmantosim, ka \(\sin ^{2} x+\cos ^{2} x=1\). Iegūstam, ka

\[\begin{gathered} (a \sin x+b \cos x)^{2} \leq\left(a^{2}+b^{2}\right)\left(\sin ^{2} x+\cos ^{2} x\right) \\ a^{2} \sin ^{2} x+2 a b \sin x \cos x+b \cos ^{2} x \leq a^{2} \sin ^{2} x+a^{2} \cos ^{2} x+b^{2} \sin ^{2} x+b^{2} \cos ^{2} x \\ 0 \leq a^{2} \cos ^{2} x-2 a b \sin x \cos x+b^{2} \sin ^{2} x \\ (a \cos x-b \sin x)^{2} \geq 0 \end{gathered}\]

Tā kā nevienādība ir patiesa jebkurai \(a, b\) un \(x\) vērtībai, tad sākotnējā nevienādība \(a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\) ir patiesa. No tā izriet, ka \(3 \sin x+4 \cos x \leq \sqrt{3^{2}+4^{2}}=5\), tātad dotajam vienādojumam nav sakņu.