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How many solutions does $$\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$$ have between $-90^\circ$ and $90^\circ$?

I used the R method and got $$2a-45^\circ = \arcsin\left(\frac{\sqrt{3}}{2}\right).$$ Since $a$ is between $-90^\circ$ and $90^\circ$, then $2a$ is between $-180^\circ$ and $180^\circ$. The RHS can be $60^\circ$, $120^\circ$, $-240^\circ$, and $-300^\circ$. Only $60^\circ$ and $120^\circ$ fit the criteria, but the answer is 4 solutions.

Where did I go wrong?

amWhy
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SuperMage1
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3 Answers3

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You are right. Here is the image of the function using google.

enter image description here

Andrei
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  • Thanks, but i want to know whats wrong with this solution that got 4 as the answer. Sin 2a - cos 2a = root 6 /2, then he squared it, getting 1-2sin2acos2a= 3/2 which is sin 4a = -1/2 which does have 4 solutions. – SuperMage1 Jan 14 '19 at 02:54
  • Since I don't see how they got 4 solutions, there is no way to explain what they did wrong. Do you know which solutions they got? – Andrei Jan 14 '19 at 02:56
  • No solutions were given – SuperMage1 Jan 14 '19 at 02:58
  • @SuperMage1 when we square we add solutions $x=2$ square $x^2 = 4$ two solutions -/+2. If you square you need to return to the original equation and check the answers – IrbidMath Jan 14 '19 at 03:25
  • @Ameryr That's why you have a plot of the original equation, so you can confirm that there are only two solutions, not 4 – Andrei Jan 14 '19 at 03:28
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Hint:

$$2a-45^\circ=180^\circ n+(-1)^n\arcsin\dfrac{\sqrt3}2$$

If $n$ is odd$=2m+1$(say)

$$2a-45=360m+180-60$$

But $-180-45\le2a-45\le180-45$

$-225\le360m+120\le135$

$?\le m\le?$

What if $n$ is even $=2m$(say)

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$\sin(2a-45) = \frac{\sqrt{3}}{2}$

$2a-45 = 60 + 360n \Rightarrow a = 52.5 + 180n$

$2a-45 = 120+ 360n \Rightarrow a = 82.5 + 180n$

Either $a = 52.5 $ or $a = 82.5$. If you square

$\sin(4a) = \frac{-1}{2}$

$4a = 210 + 360n \Rightarrow a = 52.5 + 90n $

$4a = 330 + 360n \Rightarrow a = 82.5 + 90n $ First two solution $a = 52.5 $

$a= 52.5 - 90 = -38.5$

The other two $a = 82.5 , 82.5 - 90 = -7.5$

In both cases the second solution is rejected see Checking $\sin(-15) - \cos(-15) = \frac{-\sqrt{6}}{2}$ which is wrong same for the other

IrbidMath
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