Trigonometry> Solve: sin –1 (|x 2 –1|) +cos –1 (|2x 2 –...

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

Y RAJYALAKSHMI , 10 Years ago

Grade 12th pass

2 Answers

Jitender Singh

Last Activity: 10 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

Last Activity: 10 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..