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Question: Solve the expression: \(\cos 2x + 7 = a(2 - \sin x)\) can have a real solution for A.All values o...

Solve the expression: cos⁡2x+7=a(2−sin⁡x)\cos 2x + 7 = a(2 - \sin x) can have a real solution for
A.All values of aa
B.a∈[2,6]a \in \left[ {2,6} \right]
C.a∈[−∞,2]a \in \left[ { - \infty ,2} \right]
D.a∈[0,∞]a \in \left[ {0,\infty } \right]

Explanation

Solution

Hint : In this question we have been given cos⁡2x+7=a(2−sin⁡x)\cos 2x + 7 = a(2 - \sin x) . We can see that we have trigonometric functions in this expression, so we will try to apply the trigonometric formula to simplify this.
We will use the formula
cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2{\sin ^2}\theta , by comparing here we have
θ=x\theta = x , so we can write
cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2{\sin ^2}x .

Complete step-by-step answer :
We have been given
cos⁡2x+7=a(2−sin⁡x)\cos 2x + 7 = a(2 - \sin x) .
We will put this value
cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2{\sin ^2}x in the above expression, so we have:
1−2sin⁡2x+7=2a−asin⁡x1 - 2{\sin ^2}x + 7 = 2a - a\sin x
We can transfer all the terms to the left hand side, we have:
1−2sin⁡2x+7−2a+asin⁡x=01 - 2{\sin ^2}x + 7 - 2a + a\sin x = 0
By arranging the terms we have:
−2sin⁡2x+asin⁡x+7+1−2a+=0- 2{\sin ^2}x + a\sin x + 7 + 1 - 2a + = 0
We will change the sign without changing its original form, so we have:
2sin⁡2x−asin⁡x+2a−8=02{\sin ^2}x - a\sin x + 2a - 8 = 0
Now we can see that the above expression is in the form of a quadratic form:
ax2+bx+c=0a{x^2} + bx + c = 0
We will apply the quadratic formula i.e.
x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} , we will now put the values in the formula and we have:
a=2,b=−a,c=2a−8a = 2,b = - a,c = 2a - 8
By substituting the values, we have;
sin⁡x=−(−a)±a2−4×2×(a−8)2×2\sin x = \dfrac{{ - ( - a) \pm \sqrt {{a^2} - 4 \times 2 \times (a - 8)} }}{{2 \times 2}}
Simplifying the values we have:
sin⁡x=a±a2−8×2(a−4)4\sin x = \dfrac{{a \pm \sqrt {{a^2} - 8 \times 2(a - 4)} }}{4}
We have taken the common factor out, and now by multiplying we have:
sin⁡x=a±a2−16a−644\sin x = \dfrac{{a \pm \sqrt {{a^2} - 16a - 64} }}{4}
On simplifying and removing the square root we have:
sin⁡x=a±(a−8)4\sin x = \dfrac{{a \pm (a - 8)}}{4}
We will now solve this, we can write
sin⁡x=a+(a−8)4\sin x = \dfrac{{a + (a - 8)}}{4}
It gives us
2a−84=2(a−4)4\dfrac{{2a - 8}}{4} = \dfrac{{2(a - 4)}}{4}
So it gives the value
sin⁡x=a−42\sin x = \dfrac{{a - 4}}{2}
Another value, we can take the negative sign i.e.
sin⁡x=a−(a−8)4\sin x = \dfrac{{a - (a - 8)}}{4}
On simplifying, it gives:84=2\dfrac{8}{4} = 2
We know that the value sin⁡x\sin x cannot be 22 , because the value of sin⁡x\sin xcan never exceed 11 .
So the correct value is
sin⁡x=a−42\sin x = \dfrac{{a - 4}}{2}
Now we know that the value of sine lies between −1 - 1 to 11 .
It can be written as
−1⩽sin⁡x⩽1- 1 \leqslant \sin x \leqslant 1
By putting the value we have:
−1⩽a−42⩽1- 1 \leqslant \dfrac{{a - 4}}{2} \leqslant 1
By multiplying both the sides by 22 , we have
−2⩽a−4⩽2- 2 \leqslant a - 4 \leqslant 2
We will now add 44 on both the sides of equation i.e.
4−2⩽a−4+4⩽2+44 - 2 \leqslant a - 4 + 4 \leqslant 2 + 4
It gives us value:
2⩽a⩽62 \leqslant a \leqslant 6
Therefore it gives:
a∈[2,6]a \in \left[ {2,6} \right]
Hence the correct option is (b) a∈[2,6]a \in \left[ {2,6} \right]
So, the correct answer is “Option b”.

Note : WE should note that in the above solution,
sin⁡x=a±a2−16a−644\sin x = \dfrac{{a \pm \sqrt {{a^2} - 16a - 64} }}{4} , we know that
a2−16a+64=a2−2×a×8+(8)2{a^2} - 16a + 64 = {a^2} - 2 \times a \times 8 + {(8)^2} .
So we will apply the formula i.e.
(a−b)2=a2−2ab+b2{(a - b)^2} = {a^2} - 2ab + {b^2}
So by applying this we can write
(a−8)2{(a - 8)^2} .
By putting this we have: sin⁡x=a±(a−8)24⇒a±(a−8)4\sin x = \dfrac{{a \pm \sqrt {{{(a - 8)}^2}} }}{4} \Rightarrow \dfrac{{a \pm (a - 8)}}{4}