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$\cos x \cos 2x\cos 3x= \dfrac 1 4 $

Attempt explained:

$(2\cos x \cos 3x)\cos 2x = \frac1 2 $

$(\cos 4x +\cos 2x )\cos 2x = \frac 1 2 \\\cos ^2y + \cos y (2\cos^2y- 1)= \frac1 2 \\ $ (Let, y = 2x)

$\implies 4\cos^3 y+2\cos^2y- 2\cos y-1=0$

I solved this equation using Rational Root Theorem and got $y = \frac 1 2$

$\implies x= m\pi \pm \dfrac\pi3 \forall x\in \mathbb {Z}$

Using Remainder theorem, the other solution is $\cos^2 y = \dfrac 1 2 $

$\implies x = \dfrac n2\pi \pm\dfrac \pi 8 $

But answer key states: $ x= m\pi \pm \dfrac\pi3or x =(2n+1)\dfrac \pi 8$

Why don't I get the second solution correct?

Community
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Archer
  • 6,051

4 Answers4

4

It's $$\cos2x(\cos2x+\cos4x)=\frac{1}{2}.$$ Now, let $\cos2x=t$.

Thus, $$t(t+2t^2-1)=\frac{1}{2}$$ or $$4t^3+2t^2-2t-1=0$$ or $$2t^2(2t+1)-(2t+1)=0$$ or $$(2t^2-1)(2t+1)=0,$$ which gives $$\cos4x=0$$ or $$\cos2x=-\frac{1}{2}$$ and we can write the answer.

Michael Rozenberg
  • 194,933
2

How does $\cos^2y=\dfrac12$ imply $$\dfrac{n\pi}2\pm\dfrac\pi6?$$

In fact $\cos^2y=\dfrac12\iff\cos2y=0$

$\implies2y=m\pi+\dfrac\pi2,4y=\pi(2m+1)$ where $m$ is any integer $\ \ \ \ (1)$

Again, $\cos^2y=\cos^2\dfrac\pi4\iff\sin^2y=\sin^2\dfrac\pi4$

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $

$y=n\pi\pm\dfrac\pi4,4y=\pi(4n\pm1)$ where $n$ is any integer $\ \ \ \ (2)$

Set some values of $m,n$ to identify $\{2m+1\},\{4n\pm1\}$ to be esentially the same set

lab bhattacharjee
  • 274,582
1

Remember that a cubic equation can have up to three real solutions. You did find that $\cos y = -\dfrac{1}{2}$ is one of the solutions, but you still need to find the other two.

Since $\cos y = -\dfrac{1}{2}$ is a root of $4\cos^3 y+2\cos^2y- 2\cos y-1=0$, we can factor the equation as:

\begin{align*} 4\cos^3 y+2\cos^2y- 2\cos y-1 &= 0 \\ (2\cos y + 1)(2\cos^2 y - 1) &= 0 \\ (2\cos y + 1)(\sqrt{2}\cos y + 1)(\sqrt{2}\cos y - 1) &= 0 \end{align*}

So the solutions are $$\cos y = -\dfrac{1}{2}, -\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}$$

Now, solve these for $y$, and then divide by $2$ to get $x$.

JimmyK4542
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0

All my answers are correct. However, the last answer had to be obtained using a different form. I had used the general solution of $\cos^2 x = \cos ^2 \alpha$ but they expected me to use the general form of $\cos x = 0$

"Correction":

$\cos^2 y = \frac 1 2 \implies \cos 2y = \cos 0 \implies 2y = (2n+1)\frac\pi2 \implies x = (2n+1)\dfrac \pi 8$

Any alternative methods to solve the equation are welcome.

Archer
  • 6,051