sin-1(4x)+sin-1(3x)=-π\2
Answers
Answer:
Inverse Circular Function
Formula:
sin^{-1}x+sin^{-1}y=sin^{-1}[x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}]sin
−1 x+sin −1 y=sin −1 [x 1−y 2 +y 1−x 2 ]
Now we solve the problem:
\quad sin^{-1}(4x)+sin^{-1}(3x)=-\frac{\pi}{2}sin
−1 (4x)+sin −1 (3x)=− 2π
\to sin^{-1}[4x\sqrt{1-9x^{2}}+3x\sqrt{1-16x^{2}}]=-\frac{\pi}{2}→sin
−1 [4x 1−9x 2 +3x 1−16x 2 ]=− 2π
\to 4x\sqrt{1-9x^{2}}+3x\sqrt{1-16x^{2}}=sin(-\frac{\pi}{2})=-1→4x
1−9x 2 +3x 1−16x 2 =sin(− 2π )=−1
\to 4x\sqrt{1-9x^{2}}+1=-3x\sqrt{1-16x^{2}}→4x
1−9x
2
+1=−3x 1−16x 2
Squaring both sides, we get:
\quad 16x^{2}(1-9x^{2})+8x\sqrt{1-9x^{2}}+1=9x^{2}(1-16x^{2})16x
2 (1−9x 2 )+8x 1−9x 2 +1=9x 2 (1−16x 2 )
\to 16x^{2}-144x^{4}+8x\sqrt{1-9x^{2}}+1=9x^{2}-144x^{4}→16x
2 −144x 4 +8x 1−9x 2 +1=9x 2 −144x 4
\to 8x\sqrt{1-9x^{2}}=-7x^{2}-1→8x
1−9x 2 =−7x 2 −1
Squaring both sides, we get:
\quad 64x^{2}(1-9x^{2})=49x^{4}-14x^{2}+164x
2 (1−9x 2 )=49x 4 −14x 2 +1
\to 64x^{2}-576x^{4}=49x^{4}-14x^{2}+1→64x
2 −576x 4 =49x 4 −14x 2 +1
\to 625x^{4}-50x^{2}+1=0→625x
4 −50x 2 +1=0
\to (25x^{2}-1)^{2}=0→(25x
2 −1) 2 =0
Either 25x^{2}-1=025x 2
−1=0 or, 25x^{2}-1=025x 2 −1=0
[tex]\implies \boxed{x=\pm \frac{1}{5},\:\pm \frac{1}{5}}⟹
x=±
51 ,± 51
This is the required solution.
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬