sin-3-x-cos-4-x-0- Tinku Tara June 4, 2023 Trigonometry 0 Comments FacebookTweetPin Question Number 92399 by jagoll last updated on 06/May/20 sin3(x)+cos4(x)=0 Commented by john santu last updated on 06/May/20 sin3x+(1−sin2x)2=0[sinx=p]p3+p4−2p2+1=0⇒p4+p3−2p2+1=0useFerrari′smethod Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: 1-x-1-x-2-dx-x-sin-w-cos-w-dw-sin-w-cos-w-dw-tan-w-1-sec-2-w-dw-tan-w-1-sec-2-w-du-u-1-u-2-1-u-tan-w-du-2-u-1-u-du-Next Next post: x-3-d-2-y-dx-2-x-2-dy-dx-2-ln-x- Leave a Reply Cancel replyYour email address will not be published.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.
![sin^3 x + (1−sin^2 x)^2 = 0 [ sin x=p ] p^3 +p^4 −2p^2 +1 = 0 ⇒p^4 +p^3 −2p^2 +1 = 0 use Ferrari′s method](https://www.tinkutara.com/question/Q92400.png)