Hello,

Given eqn is,

cos 2x + k ( sin x ) = 2k - 7

So, 1 - 2 sin ^ 2 x + k ( sin x ) = 2k - 7

So, 2 sin ^ 2 x - k ( sin x ) + 2k - 8 = 0

So, 2 sin ^ 2 x - k ( sin x ) + 2 ( k - 4 ) = 0............. ( 1 )

The eqn ( 1 ) is a quadratic eqn and after solving it, we get,

sin x = ( k - 4 ) / 2 and 2

SInce, value of sin x can never be greater than 1, sin x is not equal to 2

So, sin x = ( k - 4 ) / 2

Since the value of sin x ranges from -1 to 1,

-1 ≤ ( k - 4 ) / 2 ≤ 1

So, 2 ≤ k ≤ 6

So, the correct option is Option D.

Best Wishes.

Hello,

The correct answer is option d

When we solve the above equation in terms of quadratic form of sin x, then the value of k obtained solved within the range [-1,1] yields the result 2<k<6