Just a moment

     
cos²(2x)-sin(2x) = 1 -sin(2x)= 1-cos²(2x) -sin(2x)=sin²(2x) 0= sin²(2x)+sin(2x) sin(2x)(sin(2x)+1)=0 Then, you get two equations: sin(2x)=0 & sin(2x)+1=0 sin(2x)=sin (180) ...

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How bởi vì you simplify the expression displaystyle-left(sin^2x+cos^2x ight) ?
-1 is the answer to lớn the given expression.Explanation:We know, displaystyleleft(sin^2x+cos^2x=1 ight)displaystyleleft(-left(sin^2x+cos^2x=-left(1 ight) ight) ight) ...
It can be simplified todisplaystylecos^2left(2x ight)ordisplaystylefrac12left(cos4x+1 ight) Explanation:We have that displaystyle1-4sin^2xcdotcos^2x=left(1^2-left(2cosxcdotsinx ight)^2 ight)=left(1-left(sin2x ight)^2 ight)=cos^2left(2x ight) ...

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Is displaystylefleft(x ight)=cos2x-sin^2x increasing or decreasing at displaystylex=fracpi6 ?
displaystylefleft(x ight)is decreasing atdisplaystylefracpi6 Explanation:To check if this function is increasing or decreasing we should computedisplaystyleleft(f'left(fracpi6 ight) ight) ...
How vày you find all solutions of the equation in the interval displaystyleleft<0,2pi ight) given displaystylesin^2x+2cosx=2 ?
https://socratic.org/questions/how-do-you-find-all-solutions-of-the-equation-in-the-interval-0-2pi-given-sin-2x
x=0Explanation: displaystylesin^2x=1-cos^2xdisplaystyle1-cos^2x+2cosx=2 letdisplaystyleA=cosx Now solve displaystyle-A^2+2A-1=0 ...

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https://math.stackexchange.com/questions/2158247/verify-that-int-0-pi-2-f-sin-2x-cos-x-mathrm-dx-int-0-pi-2f-cos2-x
I solved it. Observe that for xin<0,pi/4> the function sin 2x is injective và positive, so the change of variable sin 2x=cos^2 t is easy lớn handle in this interval and sin 2x=cos^2 timplies sin 2x+1=(cos x+sin x)^2=cos^2 t+1 ...
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left< eginarray l l 2 & 3 \ 5 & 4 endarray ight> left< eginarray l l l 2 và 0 và 3 \ -1 & 1 & 5 endarray ight>
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