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1. Given that 3 cos 2β + 19 cos β + 13 = 0, where 90◦ < β < 180◦ , find the exact value of secB.

Factorising and solving gives me :
CosB = -2/3 and CosB=-5/2
, which is then converted to 1/cosB=-2/3 hence SecB= -3/2.

Why is the solutions CosB=-5/2 not included as the solution?
2. If you draw a cos graph you will see that the range for cosx is from 1 to -1 and hence you cannot have the -2.5 value.
3. (Original post by Chelsea12345)
Given that 3 cos 2β + 19 cos β + 13 = 0, where 90◦ < β < 180◦ , find the exact value of secB.

Factorising and solving gives me :
CosB = -2/3 and CosB=-5/2
, which is then converted to 1/cosB=-2/3 hence SecB= -3/2.

Why is the solutions CosB=-5/2 not included as the solution?
-5/2 is less than -1 so cos(β) = -5/2 is an invalid solution.

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Updated: March 11, 2018
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