MHB Emma's question at Yahoo Answers regarding solving a trigonometric equation

AI Thread Summary
The discussion revolves around solving the trigonometric equation cos²X - 8sinXcosX + 3 = 0 within the interval 0° ≤ X ≤ 360°. The initial approach involves using double-angle identities to transform the equation, leading to a more complex solution process. A simpler method is later presented, utilizing a Pythagorean identity to factor the equation, yielding two sets of solutions. The final solutions found are approximately 33.69°, 63.43°, 213.69°, and 243.43°. The exchange highlights the value of different problem-solving strategies in trigonometry.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

TRIG EQUATION, PLEASE HELP ME IM BEGGIN YOU!?

Solve the following trig equation: 0<=X<=360
cos^2 X -8sinXcosX +3=0

Please explain your answer thank you

I have posted a link there to this topic so the OP can see my work.
 
Mathematics news on Phys.org
Hello Emma,

We are given to solve:

$$\cos^2(x)-8\sin(x)\cos(x)+3=0$$ where $$0^{\circ}\le x\le360^{\circ}$$

If we use the following double-angle identities for sine and cosine:

$$\cos(2\theta)=2\cos^2(\theta)-1\,\therefore\,cos^2(\theta)=\frac{1+\cos(2\theta)}{2}$$

$$\sin(2\theta)=2\sin(\theta)\cos(\theta)$$

then the equation becomes:

$$\frac{1+\cos(2x)}{2}-4\sin(2x)+3=0$$

which we can arrange as:

$$8\sin(2x)-\cos(x)=7$$

Now, if we define (where $$k\in\mathbb{R},\,0^{\circ}<\alpha<90^{\circ}$$):

$$8=k\cos(\alpha)$$

$$1=k\sin(\alpha)$$

then we obtain by division:

$$\tan(\alpha)=\frac{1}{8}\,\therefore\,\alpha=\tan^{-1}\left(\frac{1}{8} \right)$$

and by squaring and adding:

$$8^2+1^2=65=k^2\left(cos^2(\alpha)+\sin^2(\alpha) \right)=k^2\,\therefore\,k=\sqrt{65}$$

Hence, our equation becomes:

$$\cos(\alpha)\sin(2x)-\sin(\alpha)\cos(x)=\frac{7}{k}$$

Using the angle-difference identity for sine, and substituting for $\alpha$ and $k$ we obtain:

$$\sin\left(2x-\tan^{-1}\left(\frac{1}{8} \right) \right)=\frac{7}{\sqrt{65}}$$

Because of the periodicity of the sine function, we may write, where $$k\in\mathbb{Z}$$

(1) $$\sin\left(2x-\tan^{-1}\left(\frac{1}{8} \right)+k\cdot360^{\circ} \right)=\frac{7}{\sqrt{65}}$$

Now, combining this with the identity $$\sin\left(180^{\circ}-\theta \right)=\sin(\theta)$$ we also have:

(2) $$\sin\left(\tan^{-1}\left(\frac{1}{8} \right)-2x+(2k+1)\cdot180^{\circ} \right)=\frac{7}{\sqrt{65}}$$

Taking the inverse sine of both sides of (1), we find:

$$2x-\tan^{-1}\left(\frac{1}{8} \right)+k\cdot360^{\circ}=\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)$$

Solving for $x$, we find:

$$x=\frac{1}{2}\left(\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)+\tan^{-1}\left(\frac{1}{8} \right) \right)-k\cdot180^{\circ}$$

Thus, for appropriate values of $k$, we find the solutions on the given interval for $x$ of:

$$k=0\implies x=\frac{1}{2}\left(\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)+\tan^{-1}\left(\frac{1}{8} \right) \right) \approx33.69006752598^{\circ}$$

$$k=-1\implies x=\frac{1}{2}\left(\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)+\tan^{-1}\left(\frac{1}{8} \right) \right)+180^{\circ} \approx213.69006752597983^{\circ}$$

Taking the inverse sine of both sides of (2), we find:

$$\tan^{-1}\left(\frac{1}{8} \right)-2x+(2k+1)\cdot180^{\circ}=\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)$$

Solving for $x$, we find:

$$x=\frac{1}{2}\left(\tan^{-1}\left(\frac{1}{8} \right)-\sin^{-1}\left(\frac{7}{\sqrt{65}} \right) \right)+(2k+1)\cdot90^{\circ}$$

Thus, for appropriate values of $k$, we find the solutions on the given interval for $x$ of:

$$k=0\implies x=\frac{1}{2}\left(\tan^{-1}\left(\frac{1}{8} \right)-\sin^{-1}\left(\frac{7}{\sqrt{65}} \right) \right)+90^{\circ} \approx63.43494882292201^{\circ}$$

$$k=1\implies x=\frac{1}{2}\left(\tan^{-1}\left(\frac{1}{8} \right)-\sin^{-1}\left(\frac{7}{\sqrt{65}} \right) \right)+270^{\circ} \approx243.43494882292202^{\circ}$$
 
While I was busy composing the above posts, someone else at Yahoo! Answers replied with a much simpler method, which I will outline for the benefit of our members.

We are given to solve:

$$\cos^2(x)-8\sin(x)\cos(x)+3=0$$ where $$0^{\circ}\le x\le360^{\circ}$$

Using a Pythagorean identity, we may write the equation as:

$$\cos^2(x)-8\sin(x)\cos(x)+3\left(\sin^2(x)+\cos^2(x) \right)=0$$

We may arrange this as:

$$4\cos^2(x)-8\sin(x)\cos(x)+3\sin^2(x)=0$$

Factoring, we obtain:

$$\left(2\cos(x)-\sin(x) \right)\left(2\cos(x)-3\sin(x) \right)=0$$

From the first factor, we obtain the solutions (where $$k\in\mathbb{k}$$):

$$x=\tan^{-1}(2)+k\cdot180^{\circ}$$

and for appropriate values of $k$, we obtain:

$$k=0\implies x=\tan^{-1}(2)\approx63.43494882292201^{\circ}$$

$$k=1\implies x=\tan^{-1}(2)+180^{\circ}\approx243.43494882292202^{\circ}$$

From the second factor, we obtain the solutions:

$$x=\tan^{-1}\left(\frac{2}{3} \right)+k\cdot180^{\circ}$$

and for appropriate values of $k$, we obtain:

$$k=0\implies x=\tan^{-1}\left(\frac{2}{3} \right)\approx33.69006752598^{\circ}$$

$$k=1\implies x=\tan^{-1}\left(\frac{2}{3} \right)+180^{\circ}\approx213.69006752597977^{ \circ}$$

I sure wish I had realized this method first! (Tmi)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top