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-2sin2x+cosx=0

how would i work this out?

 Sep 18, 2015
 #1
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-2sin2x+cosx=0     

 

-2(2sin xcosx) + cosx = 0    factor out cosx

 

cosx(-4sinx + 1) = 0    set each factor to 0

 

cosx  = 0     and this happens at 90° + n180°    where n is an integer

 

-4sin x + 1   = 0     subtract 1 from each side

 

-4sinx = -1    divide both sides by  -4

 

sin x = 1/4      and using the sin inverse, we have sin-1(1/4)  = x = about 14.48°  + n360°     and x = about 165.52° + n360°    where, again, n is an integer

 

Here's a graph.........https://www.desmos.com/calculator/77jadcnenw

 

 

cool cool cool

 Sep 18, 2015

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