April 21st, 2019, 03:14 AM  #1 
Newbie Joined: Apr 2019 From: Russia Posts: 3 Thanks: 0  Trigonometric equations
I am preparing for my math exam and I need your help solving these trigonometric equations: $\sin^6x + \cos^6x = \frac{7}{16}$ $\sin^2 x + \sin^2 2x = 1$ $\sin 3x + \sin 2x = \sin x$ Last edited by greg1313; April 21st, 2019 at 03:36 AM. 
April 21st, 2019, 04:24 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond 
$\displaystyle \sin^6(x)+\cos^6(x)=\frac{7}{16}$ $\displaystyle (\sin^2(x)+\cos^2(x))^33\sin^4(x)\cos^2(x)3\sin^2(x)\cos^4(x)=\frac{7}{16}$ $\displaystyle 13\sin^4(x)\cos^2(x)3\sin^2(x)\cos^4(x)=\frac{7}{16}$ $\displaystyle 3\sin^2(x)\cos^2(x)(\sin^2(x)+\cos^2(x))=\frac{9}{ 16}$ $\displaystyle \sin^2(x)\cos^2(x)=\frac{3}{16}$ $\displaystyle \sin^2(2x)=\frac34$ $\displaystyle \sin(2x)=\pm\frac{\sqrt3}{2}$ I'll leave it to you to finish up. If you get stuck post back. Last edited by greg1313; April 21st, 2019 at 09:29 AM. 
April 21st, 2019, 05:04 AM  #3 
Newbie Joined: Apr 2019 From: Russia Posts: 3 Thanks: 0  Тrigonometric equation
Okay, I solved it. Can you help me with others, please?
Last edited by skipjack; April 21st, 2019 at 06:13 AM. 
April 21st, 2019, 06:35 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
For the second equation, use the identities $\sin^2(x) = 1  \cos^2(x)$ and $\sin(2x) = 2\sin(x)\cos(x)$. For the third equation, use the identity $\sin(3x)  \sin(x) = 2\cos(2x)\sin(x)$ and then the identity $\cos(2x) + \cos(x) = 2\cos(3x/2)\cos(x/2)$. 
April 21st, 2019, 09:40 AM  #5 
Newbie Joined: Apr 2019 From: Russia Posts: 3 Thanks: 0  Trigonometric equations
" $\displaystyle \sin^2(x)\cos^2(x)=\frac{3}{16}$ $\displaystyle \sin^2(2x)=\frac34$ " Can someone explain it for me, please. I don't understand it. Last edited by Celia8241; April 21st, 2019 at 09:47 AM. 
April 21st, 2019, 09:55 AM  #6  
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587  Quote:
$2\sin{x}\cos{x} = \sin(2x) \implies 4\sin^2{x}\cos^2{x} = \sin^2(2x) \implies \sin^2{x}\cos^2{x} = \dfrac{\sin^2(2x)}{4} = \dfrac{3}{16} \implies \sin^2(2x) = \dfrac{12}{16} = \dfrac{3}{4}$  
April 21st, 2019, 02:26 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond  $\displaystyle \sin(3x)=\sin(2x+x)=\sin(2x)\cos(x)+\sin(x)\cos(2x )=2\sin(x)\cos^2(x)+\sin(x)\cos(2x)$ $\displaystyle 2\sin(x)\cos^2(x)+\sin(x)\cos(2x)+2\sin(x)\cos(x)= \sin(x)$ Note that $\sin(x)=0$ is a solution. If $\sin(x)\ne0$ then, $\displaystyle 2\cos^2(x)+\cos(2x)+2\cos(x)=1$ $\displaystyle 2\cos^2(x)+2\cos^2(x)1+2\cos(x)1=0$ $\displaystyle 4\cos^2(x)+2\cos(x)2=0$ $\displaystyle (2\cos(x)1)(2\cos(x)+2)=0$ Again, I'll leave it to you to finish up. 
April 22nd, 2019, 04:05 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203  

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