thanks for the help
how will it be then if you have one like this 4 sin2(x) + 8 cos(x) = 7 ?
If, as I assume, $\displaystyle \sin2(x) = \sin^2(x)$ then use the identity $\displaystyle \sin^2(x)+\cos^2(x) = 1$
Your equation will be $\displaystyle 4\cos^2(x)-8\cos(x)+3=0$ which does factorise but it's probably easier to use the formula to solve for cos(x). You can then find x.
i didnīt understod everything so i did it this way
4sinēx + 8cosx = 7
sinē(x) + cosē(x) = 1 <=> sinē(x)=1-cosē(x)
replace sinē(x) with 1-cosē(x)
4(1-cosē(x)) + 8cos(x) - 7 = 0 <=> 4cosē(x) + 8cos(x) - 3 = 0
replace cos(x) with t
4tē + 8t - 3 = 0
tē + 2t - 3/4 = 0
(t+1)ē = 4/4 + 3/4 = 7/4
t1 = (7/4)^0.5 - 1
t2 = -(7/4)^0.5 - 1
x1 = cos^-1(((7/4)^0.5)-1)
but is this right to ?
Moderator edit: After deleting posts and moving things around, the ordering of how the posts should appear got messed up. For some of the responses in this thread to make sense, note that this post should come before post #4 (harish21's response).
the first equation:
i cant find the worong :/
sin(5x-((4x)/3) = 1/2
t=5x-((4x)/3)
sin(t)=1/2=sin(pi/6 +n*2pi)
5x - 4pi/3 = pi/6 + n*2pi
5x = 4pi/3 + pi/6 + n*2pi
5x = 9pi/6 + n*2pi
x = (9pi/6 + n*2pi)/5
x = 3pi/10 + n * 2pi/5
i think you will have 2 solutions will the other one be x = pi - 3pi/10 + n * 2pi/5
?