Solve for x
sin-1 4x + sin-1 3x = - 2
Answers
Step-by-step explanation:
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)=−
2
π
\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
]=−
2
π
\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(−
2
π
)=−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
\implies \boxed{x=\pm \frac{1}{5},\:\pm \frac{1}{5}}⟹
x=±
5
1
,±
5
1
This is the required solution.