Solve the equation $3\sin{y}+4\cos{y}=6$ in real numbers.

Question

Solve the equation $$3\sin{y}+4\cos{y}=6$$ in real numbers.

We divide both sides of the equation by 5: $$\dfrac{3}{5}\sin{y}+\dfrac{4}{5}\cos{y}=\dfrac{6}{5}.$$ Let's choose such a narrow angle $$\alpha$$ such that $$\alpha=\dfrac{4}{5}.$$ In this case $$\cos{a}=\sqrt{1-\sin^2{\alpha}}=\sqrt{1-\left( \dfrac{4}{5} \right)^2}=\dfrac{3}{5}$$ and we can rewrite the equality as $$\sin{y}\cos{\alpha}+\cos{y}\sin{\alpha}=\dfrac{6}{5};$$ $$\sin\left(y+\alpha\right)=\dfrac{6}{5}.$$ Since the sin values do not exceed 1, this equation has no solutions. So the given equation also has no solution.

Maybe this equation can be solved in an even shorter way, for example, using the Cauchy inequality. I would be grateful for any hint on how to do this. Thanks!

FYI, the general case of what you did is shown in the Sine and cosine combination result that $a\cos x+b\sin x = c\cos(x+\varphi)$ where $c=\operatorname{sgn}(a)\sqrt{a^2+b^2}$, so it's $c=5$ in your case. — John Omielan, Apr 22 '23 at 19:52
Five misprints: $\frac44$ instead of $\frac45,$ $\alpha=$ instead of $\sin\alpha=,$ $\sin x,\cos x$ instead of $\sin y,\cos y,$ and $\cos a=$ instead of $\cos\alpha=.$ — Anne Bauval, Apr 22 '23 at 20:00

score 3 · Answer 1 · answered Apr 22 '23 at 20:08

3

Your guess was right, Cauchy-Schwarz is the key: the equation has no solution $y\in\Bbb R$ because $$|3\sin y+4\cos y|\le\sqrt{3^2+4^2}\sqrt{\sin^2y+\cos^2y}=5.$$This avoids the use of general formulas for the resolution of such equations.

answered Apr 22 '23 at 20:08

Anne Bauval

34,650

thank you very much! looks very simple) – Alice Malinova Apr 22 '23 at 22:49

score 1 · Answer 2 · answered Apr 23 '23 at 05:09

If you use the tangent half-angle substitution for solving $$3\sin({y})+4\cos({y})=k$$ you end with the quadratic $$(k+4) t^2-6 t+(k-4)=0$$ for which $$\Delta=100-4k^2$$ which is negative if $|k|>5$. So no sultion in the real domain.

But, in the complex domain, there are solutions for $k=6$ : let $y=a+ib$, expand the sine and cosine to find (forgetting the modulo) $$a=\tan ^{-1}\left(\frac{3}{4}\right)\qquad \text{and} \qquad b=\cosh ^{-1}\left(\frac{6}{5}\right)$$

Solve the equation $3\sin{y}+4\cos{y}=6$ in real numbers.

2 Answers2