Given equation
We need to solve
( 4 − 3 ) sin x − 2 3 cos 2 x = − 4 1 + 3 , x ∈ [ − 2 π , 5 π 2 ] . (4-\sqrt{3})\sin x-2\sqrt{3}\cos^2 x=-\frac{4}{1+\sqrt{3}},
\qquad x\in\left[-2\pi,\frac{5\pi}{2}\right]. ( 4 − 3 ) sin x − 2 3 cos 2 x = − 1 + 3 4 , x ∈ [ − 2 π , 2 5 π ] .
Simplify the RHS
Rationalize:
4 1 + 3 = 4 ( 1 − 3 ) 1 − 3 = − 2 ( 1 − 3 ) = 2 ( 3 − 1 ) . \frac{4}{1+\sqrt{3}}=\frac{4(1-\sqrt{3})}{1-3}=-2(1-\sqrt{3})=2(\sqrt{3}-1). 1 + 3 4 = 1 − 3 4 ( 1 − 3 ) = − 2 ( 1 − 3 ) = 2 ( 3 − 1 ) . Hence
− 4 1 + 3 = − 2 ( 3 − 1 ) = 2 − 2 3 . -\frac{4}{1+\sqrt{3}}=-2(\sqrt{3}-1)=2-2\sqrt{3}. − 1 + 3 4 = − 2 ( 3 − 1 ) = 2 − 2 3 . So the equation becomes
( 4 − 3 ) sin x − 2 3 cos 2 x = 2 − 2 3 . (4-\sqrt{3})\sin x-2\sqrt{3}\cos^2 x=2-2\sqrt{3}. ( 4 − 3 ) sin x − 2 3 cos 2 x = 2 − 2 3 .
Use cos 2 x = 1 − sin 2 x \cos^2 x=1-\sin^2 x cos 2 x = 1 − sin 2 x
( 4 − 3 ) sin x − 2 3 ( 1 − sin 2 x ) = 2 − 2 3 . (4-\sqrt{3})\sin x-2\sqrt{3}(1-\sin^2 x)=2-2\sqrt{3}. ( 4 − 3 ) sin x − 2 3 ( 1 − sin 2 x ) = 2 − 2 3 . Expand:
( 4 − 3 ) sin x − 2 3 + 2 3 sin 2 x = 2 − 2 3 . (4-\sqrt{3})\sin x-2\sqrt{3}+2\sqrt{3}\sin^2 x=2-2\sqrt{3}. ( 4 − 3 ) sin x − 2 3 + 2 3 sin 2 x = 2 − 2 3 . Bring all terms to one side:
2 3 sin 2 x + ( 4 − 3 ) sin x − 2 = 0. 2\sqrt{3}\sin^2 x+(4-\sqrt{3})\sin x-2=0. 2 3 sin 2 x + ( 4 − 3 ) sin x − 2 = 0. Let s = sin x s=\sin x s = sin x
2 3 s 2 + ( 4 − 3 ) s − 2 = 0. 2\sqrt{3}s^2+(4-\sqrt{3})s-2=0. 2 3 s 2 + ( 4 − 3 ) s − 2 = 0.
Solve the quadratic in s s s
Factor:
2 3 s 2 + ( 4 − 3 ) s − 2 = ( 2 s + 3 ) ( 3 s − 1 ) . 2\sqrt{3}s^2+(4-\sqrt{3})s-2=(2s+\sqrt{3})(\sqrt{3}s-1). 2 3 s 2 + ( 4 − 3 ) s − 2 = ( 2 s + 3 ) ( 3 s − 1 ) . Indeed,
( 2 s + 3 ) ( 3 s − 1 ) = 2 3 s 2 − 2 s + 3 s − 3 = 2 3 s 2 + s − 3 , (2s+\sqrt{3})(\sqrt{3}s-1)=2\sqrt{3}s^2-2s+3s-\sqrt{3}=2\sqrt{3}s^2+s-\sqrt{3}, ( 2 s + 3 ) ( 3 s − 1 ) = 2 3 s 2 − 2 s + 3 s − 3 = 2 3 s 2 + s − 3 , which is not the same, so let us solve directly.
Discriminant:
Δ = ( 4 − 3 ) 2 − 4 ( 2 3 ) ( − 2 ) = 16 − 8 3 + 3 + 16 3 = 19 + 8 3 = ( 4 + 3 ) 2 . \Delta=(4-\sqrt{3})^2-4(2\sqrt{3})(-2)=16-8\sqrt{3}+3+16\sqrt{3}=19+8\sqrt{3}=(4+\sqrt{3})^2. Δ = ( 4 − 3 ) 2 − 4 ( 2 3 ) ( − 2 ) = 16 − 8 3 + 3 + 16 3 = 19 + 8 3 = ( 4 + 3 ) 2 . Thus
s = − ( 4 − 3 ) ± ( 4 + 3 ) 4 3 . s=\frac{-(4-\sqrt{3})\pm(4+\sqrt{3})}{4\sqrt{3}}. s = 4 3 − ( 4 − 3 ) ± ( 4 + 3 ) . So,
s = − 4 + 3 + 4 + 3 4 3 = 2 3 4 3 = 1 2 , s=\frac{-4+\sqrt{3}+4+\sqrt{3}}{4\sqrt{3}}=\frac{2\sqrt{3}}{4\sqrt{3}}=\frac12, s = 4 3 − 4 + 3 + 4 + 3 = 4 3 2 3 = 2 1 , s = − 4 + 3 − 4 − 3 4 3 = − 8 4 3 = − 2 3 , s=\frac{-4+\sqrt{3}-4-\sqrt{3}}{4\sqrt{3}}=\frac{-8}{4\sqrt{3}}=-\frac{2}{\sqrt{3}}, s = 4 3 − 4 + 3 − 4 − 3 = 4 3 − 8 = − 3 2 , which is impossible since ∣ sin x ∣ ≤ 1 |\sin x|\le 1 ∣ sin x ∣ ≤ 1
Hence the only valid possibility is
sin x = 1 2 . \sin x=\frac12. sin x = 2 1 .
General solutions
For sin x = 1 2 \sin x=\frac12 sin x = 2 1
x = π 6 + 2 k π or x = 5 π 6 + 2 k π , x=\frac{\pi}{6}+2k\pi \quad \text{or} \quad x=\frac{5\pi}{6}+2k\pi, x = 6 π + 2 k π or x = 6 5 π + 2 k π , for integer k k k
Count solutions in [ − 2 π , 5 π 2 ] \left[-2\pi,\frac{5\pi}{2}\right] [ − 2 π , 2 5 π ]
We list them.
From x = π 6 + 2 k π x=\frac{\pi}{6}+2k\pi x = 6 π + 2 k π
k = − 1 ⇒ x = − 11 π 6 k=-1 \Rightarrow x=-\frac{11\pi}{6} k = − 1 ⇒ x = − 6 11 π k = 0 ⇒ x = π 6 k=0 \Rightarrow x=\frac{\pi}{6} k = 0 ⇒ x = 6 π k = 1 ⇒ x = 13 π 6 k=1 \Rightarrow x=\frac{13\pi}{6} k = 1 ⇒ x = 6 13 π k = 2 ⇒ x = 25 π 6 > 5 π 2 k=2 \Rightarrow x=\frac{25\pi}{6}>\frac{5\pi}{2} k = 2 ⇒ x = 6 25 π > 2 5 π
So this branch gives 3 3 3
From x = 5 π 6 + 2 k π x=\frac{5\pi}{6}+2k\pi x = 6 5 π + 2 k π
k = − 1 ⇒ x = − 7 π 6 k=-1 \Rightarrow x=-\frac{7\pi}{6} k = − 1 ⇒ x = − 6 7 π k = 0 ⇒ x = 5 π 6 k=0 \Rightarrow x=\frac{5\pi}{6} k = 0 ⇒ x = 6 5 π k = 1 ⇒ x = 17 π 6 > 5 π 2 k=1 \Rightarrow x=\frac{17\pi}{6}>\frac{5\pi}{2} k = 1 ⇒ x = 6 17 π > 2 5 π
So this branch gives 2 2 2
Total number of solutions:
3 + 2 = 5. 3+2=5. 3 + 2 = 5.
Option check
A: 4 4 4
B: 3 3 3
C: 6 6 6
D: 5 5 5
Therefore, the correct answer is D .