If sin−1⁡x2+cos−1⁡x2=5π28, then x is equal to

We have,sin−1⁡x2+cos−1⁡x2=5π28 ⇒sin−1⁡x2+π2−sin−1⁡x2=5π28⇒2sin−1⁡x2=−πsin−1⁡x−3π28=0 ⇒sin−1⁡x=π±4π24⇒sin−1⁡x=3π4 or −π4⇒x=sin⁡3π4 or sin⁡−π4∴x=12 or −12