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Solve: sin –1 (|x 2 –1|) +cos –1 (|2x 2 –5|) = pi/2

Solve: sin–1(|x2–1|) +cos–1(|2x2–5|) = pi/2

Grade:12th pass

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Hello student, please find answer to your question
sin^{-1}(|x^{2}-1|) + cos^{-1}(|2x^{2}-5|) = \frac{\pi }{2}
sin^{-1}(t) + cos^{-1}(t) = \frac{\pi }{2}
|x^{2}-1| = |2x^{2}-5|
\frac{-\sqrt{5}}{\sqrt{2}}\leq x<-1, 1\leq x<\frac{\sqrt{5}}{\sqrt{2}}
x^{2}-1 = -(2x^{2}-5)
3x^{2}=6
x^{2}=2
x = \pm \sqrt{2}
-1\leq |x^{2}-1|\leq 1
0\leq |x^{2}-1|\leq 1
-\sqrt{2} \leq x \leq \sqrt{2}
-1\leq |2x^{2}-5|\leq 1
0\leq |2x^{2}-5|\leq 1
-\sqrt{3} \leq x \leq \sqrt{3}
Final Solution:
-\sqrt{2} \leq x \leq \sqrt{2}
So the answer is
x = \pm \sqrt{2}
Sher Mohammad IIT Delhi
askIITians Faculty 174 Points
6 years ago
sin–1(|x2–1|) +cos–1(|2x2–5|) = pi/2

|x2-1|=1, gives x=0, ± sqrt(2)

2x2-5| = 1

x=+-2, +-sart(3)
take intersection and valid solution according to domain of x..

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