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1,
s i n 2 x − s i n 2 3 x + s i n 2 2 x sin^{2}x-sin^{2}3x+sin^{2}2x s i n 2 x − s i n 2 3 x + s i n 2 2 x x ϵ R x\epsilon R x ϵ R ⇔ s i n 2 2 x + ( s i n x + s i n 3 x ) ( s i n x − s i n 3 x ) = 0 \Leftrightarrow sin^{2}2x+(sinx+sin3x)(sinx-sin3x)=0 ⇔ s i n 2 2 x + ( s i n x + s i n 3 x ) ( s i n x − s i n 3 x ) = 0 ⇔ s i n 2 2 x − 2 s i n 2 x . c o s x . 2 c o s 2 x s i n x = 0 \Leftrightarrow sin^{2}2x-2sin2x.cosx.2cos2xsinx=0 ⇔ s i n 2 2 x − 2 s i n 2 x . c o s x . 2 c o s 2 x s i n x = 0 ⇔ s i n 2 2 x − s i n 2 x s i n 4 x = 0 \Leftrightarrow sin^{2}2x-sin2xsin4x=0 ⇔ s i n 2 2 x − s i n 2 x s i n 4 x = 0 ⇔ s i n 2 x ( s i n 2 x − s i n 4 x ) = 0 \Leftrightarrow sin2x(sin2x-sin4x)=0 ⇔ s i n 2 x ( s i n 2 x − s i n 4 x ) = 0 ⇔ s i n 2 x = 0 \Leftrightarrow sin2x=0 ⇔ s i n 2 x = 0
hoặc sin2x=sin4x
Đến đây tự giải được nhé
ĐK:
x ≠ k π ( k ϵ Z ) x\neq k\pi (k\epsilon Z) x = k π ( k ϵ Z ) ( 2 t a n 2 x − 4 + 2 c o t 2 x ) − 3 ( t a n x − c o t x ) + 1 = 0 (2tan^{2}x-4+2cot^{2}x)-3(tanx-cotx)+1=0 ( 2 t a n 2 x − 4 + 2 c o t 2 x ) − 3 ( t a n x − c o t x ) + 1 = 0 ( 2 t a n x − 2 c o t x ) 2 − 3 ( t a n x − c o t x ) + 1 = 0 (\sqrt{2}tanx-\sqrt{2}cotx)^{2}-3(tanx-cotx)+1=0 ( 2 t a n x − 2 c o t x ) 2 − 3 ( t a n x − c o t x ) + 1 = 0 2 ( t a n x − c o t x ) 2 − 3 ( t a n x − c o t x ) + 1 = 0 2(tanx-cotx)^{2}-3(tanx-cotx)+1=0 2 ( t a n x − c o t x ) 2 − 3 ( t a n x − c o t x ) + 1 = 0 t a n x − c o t x = 1 tanx-cotx=1 t a n x − c o t x = 1
hoặc
t a n x − c o t x = 1 2 tanx-cotx=\frac{1}{2} t a n x − c o t x = 2 1
TH1: tanx -cotx=1
⇔ − c o s 2 x = s i n x . c o s x = s i n 2 x 2 ⇔ t a n 2 x = − 2 \Leftrightarrow -cos2x=sinx.cosx=\frac{sin2x}{2} \Leftrightarrow tan2x=-2 ⇔ − c o s 2 x = s i n x . c o s x = 2 s i n 2 x ⇔ t a n 2 x = − 2
TH2:
t a n x − c o t x = 1 2 tanx-cotx=\frac{1}{2} t a n x − c o t x = 2 1 ⇔ − c o s 2 x = s i n 2 x 4 ⇔ t a n 2 x = − 4 \Leftrightarrow -cos2x=\frac{sin2x}{4}\Leftrightarrow tan2x=-4 ⇔ − c o s 2 x = 4 s i n 2 x ⇔ t a n 2 x = − 4
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
