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galina1969 [7]
3 years ago
5

solve the equation 8xsquared+2x-3=0 hence solve the equation 8cos^2y+2cosy_3=0 for the range 0degrees <y <180

Mathematics
1 answer:
Tamiku [17]3 years ago
6 0

8x^2 + 2x - 3 = 0

(4x         )(2x      )

(4x + 3)(2x - 1) = 0

Surprise: that might actually work.

4x + 3 =0

4x = - 3

x = - 3/4    That would be between 90 and 180 because the cos(x) is minus in quad 2.

2x - 1 = 0

2x = 1

x = 1/2        This one is actually only in quad 1.

cos(y) = 1/2

y = 60 degrees.

cos(y) = - 3/4

y = 138.6 degrees. You could do this by

cos(y) = 3/4

cos(y) = 41.4 degrees. Because the 3/4 is minus, the angle is in quad 2

180 - 41.4 = 138.6. Same answer.

The calculator does in the following way.

2nd F

Cos-1(

- 0.75

)

=

which will give you 138.6 immediately.

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