x-1-and-x-2-are-solutions-of-equality-cos-pix-pi-6-sin-pix-pi-6-1-2-3-0-x-12-Find-the-value-of-x-1-x-2- Tinku Tara June 3, 2023 None 0 Comments FacebookTweetPin Question Number 143964 by naka3546 last updated on 20/Jun/21 x1andx2aresolutionsofequality:cos(πx+π6)−sin(πx−π6)=123,0⩽x⩽12Findthevalueofx1+x2. Commented by Canebulok last updated on 20/Jun/21 Solution:Bysquaringbothsides,⇒cos(πx6+π6)−sin(πx6−π6)=32⇒[cos(πx6)cos(π6)−sin(πx6)sin(π6)]−[sin(πx6)cos(π6)−cos(πx6)sin(π6)]=32 Commented by Canebulok last updated on 20/Jun/21 Solution:⇒[cos(πx6)cos(π6)−sin(πx6)cos(π6)]+[cos(πx6)sin(π6)−sin(πx6)sin(π6)]=32⇒cos(π6)[cos(πx6)−sin(πx6)]+sin(π6)[cos(πx6)−sin(πx6)]=32⇒[cos(π6)+sin(π6)][cos(πx6)−sin(πx6)]=32Bysquaringbothsides,⇒[1+2cos(π6)sin(π6)][1−2sin(πx6)cos(πx6)]=34⇒[1+sin(π3)][1−sin(πx3)]=34⇒1−sin(πx3)=34[1+sin(π3)]⇒−sin(πx3)=3(4+4sin(π3))−1⇒sin(πx3)=1−3(4+4sin(π3))⇒arcsin[1−3(4+4sin(π3))]=πx3⇒arcsin[1−3(4+4sin(π3))](3π)=x∼Kevin Answered by bramlexs22 last updated on 20/Jun/21 sin(πx−π6)=cos(3π−πx+π6)=cos(πx−4π6)⇔cos(πx+π6)−cos(πx−4π6)=32⇔−2sin(2πx−3π12)sin(5π12)=32⇔ Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: I-pi-6-pi-3-cos-6-x-1-3sin-2-xcos-2-x-dx-Next Next post: Question-12895 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.