Harder 3 Unit Questions (1 Viewer)

[MC Squidge]

i) Solve cos3@-cos@=2sin@ for 0<@<2Pi
ii) Solve 2|x|-|x-2|>2

 

[alakazimmy]

i) Solve cos3@-cos@=2sin@ for 0<@<2Pi -

cos(2@ + @) - cos(2@ - @) = 2 sin@
(cos2@cos@ - sin2@sin@) - (cos2@cos@ + sin2@sin@) = 2sin@
-2(sin2@sin@) = 2sin@
sin@ (1 + sin2@) = 0

Hence, solve for sin@ = 0, and sin2@ = -1 in for 0<@<2Pi

For sin@ = 0, @ = 0, Pi
For sin2@ = -1, 2@ = 3Pi/2, 7Pi/2

Therefore, @ = 0, 3Pi/4, Pi and 7Pi/4


ii) Solve 2|x|-|x-2|>2

For this question, you need to break it down into 3 parts.

1) For values of x < 0
2) For values of 0<x<2
3) For values of x>2


For x<0
2|x| - |x - 2| > 2
-2x + x - 2 > 2
-x - 4 > 0
Hence, x < -4

For 0<x<2
2|x| - |x - 2| > 2
2x + x - 2 > 2
3x > 4
Hence, x > ⁴/₃, but x is also less than 2
So, ⁴/₃ < x < 2

Finally, for x>2
2|x| - |x - 2| > 2
2x - x + 2 > 2
x > 0
hence, x > 2

 

[Drongoski]

i) Solve cos3@-cos@=2sin@ for 0<@<2Pi
ii) Solve 2|x|-|x-2|>2

(ii) for x > 2: LHS = 2x -x + 2 > 2 => x > 0 ==> x > 2

for 0 < x < 2: LHS = 2x - (2-x) > 2 ==> 3x > 4 ==> 4/3 < x < 2

for x < 0: LHS = -2x + x - 2 > 2 ==> -x > 4 ==> x < -4

 

[Michaelmoo]

i) Solve cos3@-cos@=2sin@ for 0<@<2Pi
ii) Solve 2|x|-|x-2|>2

For the first one, use the sums to products

cos3@-cos@ = 2sin@

2sin(3@ + @/2)sin(@-3@/2) = 2sin@

-2sin2@sin@ =2sin@

2sin2@sin@ +2sin@ = 0

2sin@(sin2@ +1) = 0

and just solve

2sin@ = 0

sin2@ + 1 = 0

 

[malady]

a useful approach for quests like (ii) is to graph y=2|x| and y= |x-2|+ 2 on the same diagram and solve 2|x| > |x-2|+2

 

[dongerluo]

a useful approach for quests like (ii) is to graph y=2|x| and y= |x-2|+ 2 on the same diagram and solve 2|x| > |x-2|+2

yep, its easier if u draw a graph for (ii)

 

[Trebla]

The graphing method is only quicker if you are lucky enough to get integer values for their intersections. This is not the case for the given question and to find that point of intersection, you have to solve 2|x| = |x - 2| + 2 algebraically anyway.

 

[Drongoski]

The graphing method is only quicker if you are lucky enough to get integer values for their intersections. This is not the case for the given question and to find that point of intersection, you have to solve 2|x| = |x - 2| + 2 algebraically anyway.

Great observation, Trebla!

 

[MC Squidge]

thanks guys i understand part 1 now

but i think u all got part 2 wrong

Function Grapher Online

graph it there

 

[Timothy.Siu]

The graphing method is only quicker if you are lucky enough to get integer values for their intersections. This is not the case for the given question and to find that point of intersection, you have to solve 2|x| = |x - 2| + 2 algebraically anyway.

yeah true, but doing a quick sketch cant be any harm, it cud make sure u that ur final answer is correct

 

[Drongoski]

thanks guys i understand part 1 now

but i think u all got part 2 wrong

Function Grapher Online

graph it there

What's the correct answer then ??

 

[Trebla]

2|x| - |x - 2| > 2

Note that:
|x|
= x when x > 0
= 0 when x = 0
= - x when x < 0
|x - 2|
= x - 2 when x > 2
= 0 when x = 2
= - x + 2 when x < 2

For x < 0:
2(- x) - (- x + 2) > 2
- x - 2 > 2
=> x < - 4
Since x < 0, then the total solution in this domain is x < - 4

For 0 ≤ x ≤ 2:
2x - (- x + 2) > 2
3x - 2 > 2
=> x > 4/3
But 0 ≤ x ≤ 2, hence total solution in this domain is 4/3 < x ≤ 2

For x > 2
2x - (x + 2) > 2
x - 2 > 2
=> x > 4
Since x > 2, then total solution is x > 4

Combining the solutions together gives x < - 4 or x > 4/3

 

[MC Squidge]

^
yea this guy got it right

 

[malady]

The graphing method is only quicker if you are lucky enough to get integer values for their intersections. This is not the case for the given question and to find that point of intersection, you have to solve 2|x| = |x - 2| + 2 algebraically anyway.

Hey Trebla, I don't think solving solving y=-2x, y= 4-x and y=2x, y=4-x to get the points of intersection is exactly rocket science <G>

 

[Trebla]

Hey Trebla, I don't think solving solving y=-2x, y= 4-x and y=2x, y=4-x to get the points of intersection is exactly rocket science <G>

My point was that you have to solve it algebraically in terms of the equality when you graph it to determine points of intersection, which is basically like solving the original inequality algebraically anyway except without the graphing.