9
$\begingroup$

Question: Solve the trigonometric equation: $\sin x + \cos x=\sin 2x + \cos 2x$.

My attempt:

$\sin x + \cos x=\sin 2x + \cos 2x$

$\implies \sin x + \cos x=2\sin x \cos x + \cos^2 x - \sin^2 x$

$\implies \sin x + \cos x=2\sin x \cos x + \cos^2 x - (1-\cos^2 x)$

$\implies \sin x + \cos x=2\sin x \cos x + 2\cos^2 x - 1$

$\implies \sin x - 2\sin x \cos x + \cos x - 2\cos^2 x= - 1$

$\implies \sin x(1-2\cos x)+\cos x(1-2\cos x)=-1$

$\implies (1-2\cos x)(\sin x+\cos x)=-1$

$\implies (1-2\cos x)=-1$ or $(\sin x +\cos x)=-1$

$\implies \cos x=1$ or $\sin^2 x +\cos^2 x + 2\sin x\cos x=1$

$\implies x=2n\pi$ or $\sin 2x=0$

$\implies x=2n\pi$ or $2x=n\pi$

$\therefore x=2n\pi$ or $x=\frac{n\pi}{2}$

But the answers given in my book are $x=2n\pi$ and $x=\frac{(4n+1)\pi}{6}$. Where have I gone wrong? Please help.

$\endgroup$
5
  • 3
    $\begingroup$ You did an error there: $(1−2cosx)=−1⟹(1−2cos⁡x)=−1 or (sinx+cosx)=−1$ $\endgroup$
    – Olimjon
    Commented Oct 18, 2017 at 4:55
  • $\begingroup$ Hint: consider $\cos(x-\frac{\pi}{4})$ $\endgroup$ Commented Oct 18, 2017 at 4:58
  • $\begingroup$ Why is my method not working? $\endgroup$
    – MrAP
    Commented Oct 18, 2017 at 4:59
  • 16
    $\begingroup$ Because $a b = -1$ doesn't imply $a=-1$ or $b = -1$. $\endgroup$ Commented Oct 18, 2017 at 5:00
  • 1
    $\begingroup$ Hint: $$\cos\left(x-\dfrac\pi4\right)=\cos\left(2x-\dfrac\pi4\right)$$ Or $$\sin(2x+\pi/4)=\sin(x+\pi/4)$$ $\endgroup$ Commented Oct 18, 2017 at 6:09

2 Answers 2

39
$\begingroup$

To expand on @gribouillis 's comment, the error in your argument is this step:

$(1-2\cos x)(\sin x+\cos x)=-1$

$\implies (1-2\cos x)=-1$ or $(\sin x +\cos x)=-1$

This is an incorrect implication.

$ab=c$ only implies $a=c$ or $b=c$ when $c=0$.

For $c=-1$ as in this case, you could have $a=1,b=-1$ or $a=5,b=-0.2$ or $a=-1000,b=0.001$ or an infinite number of other combinations.

$\endgroup$
1
  • 17
    $\begingroup$ +1 for actually answering the question "where have I gone wrong". The OP didn't ask for a solution, but an explanation of why their method failed. $\endgroup$ Commented Oct 18, 2017 at 7:18
5
$\begingroup$

Use Subtraction: $$\sin 2x−\sin x=\cos x−\cos 2x$$ $$2\sin\frac{x}2\cos\frac{3x}{2}=2\sin\frac{3x}{2}\sin\frac{x}{2}$$ So, $$\sin\frac{x}{2}=0$$ OR $$\tan\frac{3x}{2}=1$$

$\endgroup$
1
  • 12
    $\begingroup$ A good answer to the question "how to solve this", but the question here was "where is the mistake in the proposed solution". $\endgroup$
    – user193810
    Commented Oct 18, 2017 at 8:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .