Alevel (module 2/3 standard). Any help at all would be hugely appriciated.
32. a) Solve in radians
sin (x + pi/6) = 2cosx
giving all solutions in range of 0 < x < 2pi
b) Use the formular tan (a - b) = (tana - tanb)/(1 + tana tanb)
to show that tan(x - pi/4) = (tanx - 1)/(tan x+1)
c) Hence or otherwise solve in radians tan(x - pi/4) = 6tanx
67. Betweem -180 < x < 180 solve
cosec^2 x = 1 + 2 sec^2 x
68. For all values of x prove that
8 sin^4 x = 3 - 4 cos 2 x + cos 4 x
Hence find in radians in terms of pi the values of x in pi and -pi for which cos 4x - 4cos 2x + 1 = 0
70. Find the values of x for 0-360 for which tan^2 x + 2 sec x = 5
85. Find values to which
3 cos 2x + 7 cos x = 0 and 0 < x < 2pi
Thank you for reading. xx
usually i'd tell you to start working on the right, but maybe it's better to start on the left here
note,
now expand and simplify to get the right hand side
(remember your formulas for the double angle of cosine)
in light of the above, we haveHence find in radians in terms of pi the values of x in pi and -pi for which cos 4x - 4cos 2x + 1 = 0
can you continue?
32 was fine thanks.
67. [tex]
1/(sin^2x) = 1 + 2/(cos^2x)
After this, I got a bit stuck. Another hint would be useful please.
68.
Quote:
Hence find in radians in terms of pi the values of x in pi and -pi for which cos 4x - 4cos 2x + 1 = 0
in light of the above, we have
can you continue?
In short, no. I'm not 100% sure where you got the first line from, and from there how to solve the final equation.
Thanks for the help so far.
that's where i left off. you did nothing after that? do you know how to combine fractions?
you would have:
do you think you can continue now?
remember, at this point we showed that68.
Quote:
Hence find in radians in terms of pi the values of x in pi and -pi for which cos 4x - 4cos 2x + 1 = 0
in light of the above, we have
can you continue?
In short, no. I'm not 100% sure where you got the first line from, and from there how to solve the final equation.
Thanks for the help so far.
(i had a typo before, i thought it was - cos (4x) and + 4cos(2x))
add 1 to both sides, we get:
now we are told that
thus we have:
now continue
i do not recommend dividing by cos^2 x. if a solution of x ends up being one that causes cos(x) = 0, it would mean you have divided by 0, hence making your solution invalid. bring everything to one side and factor out the cos^2 x instead
no factorization necessary, just solve for sin^4 x and take it from there68. How do you simplify/factorise this?
you could have taken out cos^2 x out of the terms where it's common. but i made a post saying why it is ok to divide by cos^2 x, so don't worry about it
EDIT: i'm kind of busy with other stuff right now, and as a result, i cannot concentrate on your problems as much as i would like to. i'm going to log off. hopefully someone else will help you. if not, i'll pick up where i left off when i come back on again