J jay47 Think Positively! 5+ Year Member 15+ Year Member Mar 23, 2009 #1 Advertisement - Members don't see this ad Hey math lovers, doing some engineering with my dad for a project and need some help with this equation. 12= sin[90-(x/2)]tan[x/2] Please solve for X. I'll check your answer with mine.
Advertisement - Members don't see this ad Hey math lovers, doing some engineering with my dad for a project and need some help with this equation. 12= sin[90-(x/2)]tan[x/2] Please solve for X. I'll check your answer with mine.
Marmar tala Full Member 10+ Year Member Mar 23, 2009 #2 12=sin[90-(x/2)] tan [x/2] , sin [90-(x/2)]=-cos (x/2) ==> 12= -cos (x/2) tan [x/2]= -sin (x/2) no answer for x. always -1<sin a<1 Upvote 0 Downvote
12=sin[90-(x/2)] tan [x/2] , sin [90-(x/2)]=-cos (x/2) ==> 12= -cos (x/2) tan [x/2]= -sin (x/2) no answer for x. always -1<sin a<1
S Streetwolf Ultra Senior Member Verified Member 10+ Year Member Dentist 15+ Year Member Mar 23, 2009 #3 Marmar tala said: 12=sin[90-(x/2)] tan [x/2] , sin [90-(x/2)]=-cos (x/2) ==> 12= -cos (x/2) tan [x/2]= -sin (x/2) no answer for x. always -1<sin a<1 Click to expand... Agreed except I get sin(x/2) = 12, not -sin. Upvote 0 Downvote
Marmar tala said: 12=sin[90-(x/2)] tan [x/2] , sin [90-(x/2)]=-cos (x/2) ==> 12= -cos (x/2) tan [x/2]= -sin (x/2) no answer for x. always -1<sin a<1 Click to expand... Agreed except I get sin(x/2) = 12, not -sin.
joonkimdds Senior Member 10+ Year Member 15+ Year Member Mar 27, 2009 #4 Marmar tala said: 12=sin[90-(x/2)] tan [x/2] , sin [90-(x/2)]=-cos (x/2) ==> 12= -cos (x/2) tan [x/2]= -sin (x/2) no answer for x. always -1<sin a<1 Click to expand... what??? what's the "," comma on the first line and where did 12 go? Upvote 0 Downvote
Marmar tala said: 12=sin[90-(x/2)] tan [x/2] , sin [90-(x/2)]=-cos (x/2) ==> 12= -cos (x/2) tan [x/2]= -sin (x/2) no answer for x. always -1<sin a<1 Click to expand... what??? what's the "," comma on the first line and where did 12 go?