Given that 3sin^2 x − 8cos x − 7 = 0,

show that, for real values of x,

cos x = −2/3.

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`3sin^2 x - 8 cos x - 7 = 0`

We know that `sin^2 x + cos^2 x = 1`

`3(1-cos^2 x) -8 cos x-7 = 0`

`3-3cos^2 x-8cosx-7 = 0`

`3cos^2 x+8cosx+4 = 0`

`3cos^2 x + 6cosx+2cosx+4=0`

`3cosx(cosx+2)+2(cosx+2)=0`

`(cosx+2)(3cosx+2)=0`

`Cosx = -2` or `cosx = -2/3`

We know that for real values of x `-1<= cosx<=1`

`cosx= - 2/3`

*So the required answer is proved.*

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