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core 4 solving trig

need some help on this question, thanks.

Given that 0º < x < 360º,
(i) solve 2cos x – sin x = l,
(ii) deduce the solution set of the inequality
2cos x – sin x > 1.

for (i) i got cosx=2/3 so x=48.2 and 228.2, but it seems wrong, so can any1 see what i did wrong?

for (ii) i comepletely forgot how to do that.
Reply 1
Avatar for Slowbro93
Slowbro93
20
Original post by zero334
need some help on this question, thanks.

Given that 0º &lt; x &lt; 360º,
(i) solve 2cos x – sin x = l,
(ii) deduce the solution set of the inequality
2cos x – sin x &gt; 1.

for (i) i got cosx=2/3 so x=48.2 and 228.2, but it seems wrong, so can any1 see what i did wrong?

for (ii) i comepletely forgot how to do that.


For i, you could use Rcos(x+alpha) and then solve for x

For ii, use a graph once you have your new equation. Then use this information to work out this inequality.
Reply 2
Avatar for ztibor
ztibor
10
Original post by zero334
need some help on this question, thanks.

Given that 0º < x < 360º,
(i) solve 2cos x – sin x = l,
(ii) deduce the solution set of the inequality
2cos x – sin x > 1.

for (i) i got cosx=2/3 so x=48.2 and 228.2, but it seems wrong, so can any1 see what i did wrong?

for (ii) i comepletely forgot how to do that.


For (i)
arrange sinx to the RHS and take square of both side
4cos^2(x)= 1+2sinx+sin^2(x)
use that cos^2(x)=1-sin^2(x) this gives a quadratic equation for sinx

For(ii)
Use the first quadratic equation to solve the inequality
(edited 13 years ago)
Reply 3
Avatar for Potassium^2
Potassium^2
If you square both sides, always check you don't produce spurious solutions at the end
Original post by Potassium^2
If you square both sides, always check you don't produce spurious solutions at the end

Thank God someone said it!

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