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Trig Equation

cos⁡2(x)=2cos⁡(x)sin⁡(x)−0.6\cos^2(x)=2\cos(x)\sin(x)-0.6. Solve for x.
Reply 1
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mqb2766
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Original post by esrever
cos⁡2(x)=2cos⁡(x)sin⁡(x)−0.6\cos^2(x)=2\cos(x)\sin(x)-0.6. Solve for x.


double angle formula?
Reply 2
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8013
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cos^2(x)-2cos(x)sin(x)+0.6=0
sin^2(x)=0.6
Reply 3
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esrever
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20
Original post by mqb2766
double angle formula?


Do you mean rewriting 2sin(x)cos(x) as sin(2x)?
Reply 4
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esrever
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Original post by 8013
cos^2(x)-2cos(x)sin(x)+0.6=0
sin^2(x)=0.6


Can you please provide some more steps?
Reply 5
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mqb2766
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Original post by esrever
Do you mean rewriting 2sin(x)cos(x) as sin(2x)?


yes and similar for cos^2(x)
Reply 6
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esrever
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Original post by mqb2766
yes and similar for cos^2(x)


I tried it but got cos⁡(2x)=2sin⁡(2x)−2.2\cos(2x) = 2\sin(2x) - 2.2. I don't know where to proceed from here.
(edited 5 years ago)
Reply 7
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mqb2766
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Good point :-). Give me a sec.
Original post by esrever
I tried it but got cos⁡(2x)=2sin⁡(2x)−2.2\cos(2x) = 2\sin(2x) - 2.2. I don't know where to proceed from here.
Reply 8
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mqb2766
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Original post by mqb2766
Good point :-). Give me a sec.


Convert the cos and sin into a single
Rsin(2x + alpha)
then solve.
Original post by 8013
cos^2(x)-2cos(x)sin(x)+0.6=0
sin^2(x)=0.6


{cos(2x) + 1}/2 -sin(2x) + 0.6 = 0

cos(2x) - 2sin(2x) = -2,2

now apply the wave form method.... Rcos(2x + α) = -2.2
Reply 10
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esrever
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Thank you so much :smile:. But my textbook quotes an alternative method (which I found quite counter-intuitive):

Rewrite as it as 1=2tan⁡(x)−0.6sec⁡2(x)1=2\tan(x)-0.6\sec^2(x) and then use tan⁡2(x)+1=sec⁡2(x)\tan^2(x) + 1 = \sec^2(x).

Isn't there a easier way similar to this?
Original post by esrever
Thank you so much :smile:. But my textbook quotes an alternative method (which I found quite counter-intuitive):

Rewrite as it as 1=2tan⁡(x)−0.6sec⁡2(x)1=2\tan(x)-0.6\sec^2(x) and then use tan⁡2(x)+1=sec⁡2(x)\tan^2(x) + 1 = \sec^2(x).

Isn't there a easier way similar to this?


When you have a quadratic like this, the usual ways are to 1) convert into a double angle (as above) or 2) solve as a quadratic (as textbook). It looked like a double angle because of the cos^2 and cos*sin terms, but there are usually a few different ways to solve and as long as you spot the sub, the textbook answer is ok.
(edited 5 years ago)
Reply 12
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esrever
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Original post by mqb2766
When you have a quadratic like this, the usual ways are to 1) convert into a double angle (as above) or 2) solve as a quadratic (as textbook). It looked like a double angle because of the cos^2 and cos*sin terms, but there are usually a few different ways to solve and as long as you spot the sub, the textbook answer is ok.


Thank you so much :smile:
Original post by esrever
Thank you so much :smile:. But my textbook quotes an alternative method (which I found quite counter-intuitive):

Rewrite as it as 1=2tan⁡(x)−0.6sec⁡2(x)1=2\tan(x)-0.6\sec^2(x) and then use tan⁡2(x)+1=sec⁡2(x)\tan^2(x) + 1 = \sec^2(x).

Isn't there a easier way similar to this?


that method works. in the exam they would give you a hint... they would not expect you to spot that right away.
Reply 14
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esrever
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Original post by the bear
that method works. in the exam they would give you a hint... they would not expect you to spot that right away.


That's likely but recent further maths questions have been very tricky though

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