-2
$\begingroup$

Question

Find all $x$ that satisfy $ 3 \cos{2x}=-1 $ for $0^{\circ} \leq x \leq 180^{\circ}$

can you help to solve this? tq

$\endgroup$
2
2
$\begingroup$

$$2\cos 4x+3 \cos 2x=-1 \space \space \space \space \space \text{using} \space \space \space (\cos 4x=2\cos^2 2x-1)$$ $$4\cos^2 2x-2+3\cos 2x=-1$$ $$4\cos^2 2x+3\cos 2x-1=0$$ $$\cos 2x=\dfrac{-3\pm\sqrt{3^2-4(4)(-1)}}{2\times4}$$ $$\cos 2x=\dfrac{1}{4},-1$$ Consider the graph of $\cos 2x$ or solving one-by-one gives, $$\cos 2x=-1\Rightarrow 2x=\pi\rightarrow x=\dfrac{\pi}{2}$$ We won't consider the other solutions as it is not mentioned in the range above. $$\cos 2x=\dfrac{1}{4}$$ $$2\cos^2 x-1=\dfrac{1}{4}$$ $$\cos^2 x=\dfrac{5}{8}\Rightarrow \cos x=\pm \sqrt {\dfrac{5}{8}}$$ In the interval $[0, \pi]$ $\cos x $ takes all the values between $[-1,1]$ and so both values will occur for any $x$ in the interval or you can also check by graph. enter image description here enter image description here

$\endgroup$
7
  • 2
    $\begingroup$ please don't answer questions that have no input from the user's side. we want people to learn, hints are better. $\endgroup$
    – Vega
    Oct 10 at 4:31
  • 1
    $\begingroup$ Sorry, will remember @Vega $\endgroup$
    – PCMSE
    Oct 10 at 4:41
  • 1
    $\begingroup$ no problem :) , it was just a heads up, same happened to me once. $\endgroup$
    – Vega
    Oct 10 at 4:48
  • $\begingroup$ Thank you @Vega $\endgroup$
    – PCMSE
    Oct 10 at 4:49
  • $\begingroup$ off topic, are you a JEE aspirant? $\endgroup$
    – Vega
    Oct 10 at 4:50
0
$\begingroup$

Hint: Use $\cos(2\theta) = 2\cos^2\theta-1$ and form a quadratic in terms of $\cos(2x)$. Find the value of $\cos(2x)$ and apply the identity again. Finally solve for $\cos x$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.