4

Solve $\cos x+8\sin x-7=0$

My attempt:

\begin{align} &8\sin x=7-\cos x\\ &\implies 8\cdot \left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)=7-\cos x\\ &\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=7-1+2\sin ^2\frac{x}{2}\\ &\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=6+2\sin^2 \frac{x}{2}\\ &\implies 8\sin \frac{x}{2}\cos \frac{x}{2}=3+\sin^2 \frac{x}{2}\\ &\implies 0=\sin^2 \frac{x}{2}-8\sin \frac{x}{2}\cos \frac{x}{2}+3\\ &\implies 0=\sin \frac{x}{2}\left(\sin \frac{x}{2}-8\cos \frac{x}{2}\right)+3 \end{align}

I'm not sure how to proceed from here (if this process is even right at all?) . Any help would be appreciated. Thanks.

Sujaan Kunalan
  • 10,944
  • Use http://math.stackexchange.com/questions/618192/how-to-solve-sin-x-cos-x-1/618767#618767 or http://math.stackexchange.com/questions/618192/how-to-solve-sin-x-cos-x-1/618430#618430 – lab bhattacharjee May 01 '14 at 04:32
  • We can always bring everything but $\cos x$ to one side, square, replace $\cos^2 x$ by $1-\sin^2 x$. Then we need to throw away extraneous roots. – André Nicolas May 01 '14 at 04:46

4 Answers4

6

Use the formula for linear combinations of sine and cosine:

$$A \cos x+B \sin x=C \sin (x+\phi) $$ where $C=\sqrt{A^2+B^2},\phi=\arg (A+Bi)$.

This makes transforms your equation to:

$$\sqrt{65} \sin(x+\tan^{-1} 8)=7 $$ which can be easily solved.

user1337
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  • If you were to memorize an equation, I would suggest the other form of this, $A_1 \angle \phi_1 + A_2 \angle \phi_2 = A_3 \angle \phi_3 \quad \implies \quad A_1\cos(x + \phi_1) + A_2\cos(x + \phi_2) \equiv A_3 \cos(x + \phi_3) $. – DanielV May 01 '14 at 05:20
  • @DanielIV: Memorize! I have to re-derive this stuff every time! Corroded curcuits, I guess! ;-) – Robert Lewis May 01 '14 at 05:45
6

Divide either sides by $\cos^2\frac x2$ $$\sin^2\frac x2-8\sin\frac x2\cos\frac x2+3=0$$ to get $$\tan^2\frac x2-8\tan\frac x2+3\left(1+\tan^2\frac x2\right)=0$$ which on rearrangement is a Quadratic Equation in $\displaystyle\tan\frac x2$

lab bhattacharjee
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2

Another solution could use the tangent half-angle substitution. If you define $t=\tan \frac{x}{2}$, you have $\sin x=\frac{2t}{1+t^2}$ and $\cos x=\frac{1-t^2}{1+t^2}$. So, the equation becomes $$\cos x+8\sin x-7=\frac{16 t+2}{t^2+1}-8=0$$ that is to say $$-8 t^2+16 t-6=0$$ the roots of which being $t_1=\frac{1}{2}$ and $t_2=\frac{3}{2}$ and so $x_1=2 \tan ^{-1}\left(\frac{1}{2}\right)$ and $x_2=2 \tan ^{-1}\left(\frac{3}{2}\right)$.

I am sure that you can take from here.

Claude Leibovici
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1

$1.) \; \cos x+8\sin x-7=0; \tag{1}$

$2.) \; \cos^2 x = (7- 8\sin x)^2; \tag{2}$

$3.) \; \cos^2 x = 1 - \sin^2 x; \tag{3}$

$4.) \; 1 - \sin^2 x = (7- 8\sin x)^2; \tag{4}$

$5.) \; 1 - \sin^2 x = 49 - 112 \sin x + 64 \sin^2 x; \tag{5}$

$6.) \; 65 \sin^2 x -112\sin x + 48 = 0; \tag{6}$

$7.) \; \sin x = \dfrac{112 \pm \sqrt{(112)^2 - 4(48)(65)}}{130}; \tag{7}$

$8.) \; \sin x = \dfrac{112 \pm \sqrt{12,544 -12,480}}{130}; \tag{8}$

$9.) \sin x = \dfrac{112 \pm 8}{130} = \dfrac{104}{130}, \dfrac{120}{130} = \dfrac{4}{5}, \dfrac{12}{13}; \tag{9}$

$10.) \; \text{Note there exist unique} \; \alpha, \beta \in [0, \dfrac{\pi}{2}] \; \text{with} \; \sin \alpha = \dfrac{4}{5}, \sin \beta = \dfrac{12}{13}; \tag{10}$

$11.) \; \text{Note} \; \cos \alpha = \dfrac{3}{5}, \cos \beta = \dfrac{5}{13}; \tag{11}$

$12.) \; \text{Note} \; \alpha \; \text{alone satisfies (1)}; \tag{12}$

$13.) \; \text{Note there exist unique} \; \gamma, \delta \in [\dfrac{\pi}{2}, \pi] \; \text{with} \; \sin \gamma = \dfrac{4}{5}, \sin \delta = \dfrac{12}{13}; \tag{13}$

$14.) \; \text{Note that} \; \cos \gamma = -\dfrac{3}{5}, \cos \delta = -\dfrac{5}{13}; \tag{14}$

$15.) \; \text{Note that} \; \delta \; \text{alone satisfies (1)}; \tag{15}$

$16.) \; \text{Note by periodicity} \; \alpha \pm 2n\pi, \delta \pm 2n\pi \; \text{satisfy (1) for} \; n \in \Bbb Z; \tag{16}$

$17.) \; \text{Conclude that all solutions are given by item (16) above}; \tag{17}$

$18.) \; \text{Kick back for a minute; you deserve it!} \tag{18}$

Hope this helps! Cheerio,

and as always,

Fiat Lux!!!

Robert Lewis
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