1.

5cos^2x-7sin^2x-6=0

Answer»

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

5cos^2x -7sin^2x-6 = 0

5(1-sin^2x) - 7sin^2x-6=0

5- 12sin^2x -6 = 0

1+12sin^2x= 0

sin^2x =-1/12

since, range of sin^2x =[0,1]

which means no real value of x exists as solution.


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