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core 4 solving trig watch

1. 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.
2. (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.
3. (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
4. If you square both sides, always check you don't produce spurious solutions at the end
5. (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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