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Solve 5sin^2-3cosx=4 DOMAIN=(0,2pi)

 Sep 30, 2014

Best Answer 

 #1
avatar+17746 
+5

In  5sin²x -3cosx = 4      either change the sin into cos or the cos into sin.

If you use the Pythagorean identity    sin²x  +  cos²x  = 1,

    you can get   sin²x  =  1  -  cos²x 

Substituting that into the original equation:

      5sin²x -3cosx = 4    --->   5( 1 - cos²x ) - 3 cos x  =  4

      Multiply out              --->   5 - 5 cos²x  - 3 cos x  =  4

      Write as a quadratic  --->  -5 cos²x  -  3 cos x  +  1  =  0

       Simplify                  --->    5 cos²x  +  3 cos x  -  1  =  0

Use the quadratic formula to get values for cos x; then use invcos x to get solutions.

 Sep 30, 2014
 #1
avatar+17746 
+5
Best Answer

In  5sin²x -3cosx = 4      either change the sin into cos or the cos into sin.

If you use the Pythagorean identity    sin²x  +  cos²x  = 1,

    you can get   sin²x  =  1  -  cos²x 

Substituting that into the original equation:

      5sin²x -3cosx = 4    --->   5( 1 - cos²x ) - 3 cos x  =  4

      Multiply out              --->   5 - 5 cos²x  - 3 cos x  =  4

      Write as a quadratic  --->  -5 cos²x  -  3 cos x  +  1  =  0

       Simplify                  --->    5 cos²x  +  3 cos x  -  1  =  0

Use the quadratic formula to get values for cos x; then use invcos x to get solutions.

geno3141 Sep 30, 2014
 #2
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0

Thanks

 Oct 1, 2014

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