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

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

Guest Sep 30, 2014

#1
+17721
+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.

geno3141  Sep 30, 2014
Sort:

#1
+17721
+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.

geno3141  Sep 30, 2014
#2
0

Thanks

Guest Oct 1, 2014

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