esrever
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#1
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\cos^2(x)=2\cos(x)\sin(x)-0.6. Solve for x.
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mqb2766
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(Original post by esrever)
\cos^2(x)=2\cos(x)\sin(x)-0.6. Solve for x.
double angle formula?
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8013
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cos^2(x)-2cos(x)sin(x)+0.6=0
sin^2(x)=0.6
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esrever
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(Original post by mqb2766)
double angle formula?
Do you mean rewriting 2sin(x)cos(x) as sin(2x)?
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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?
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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)
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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) = 2\sin(2x) - 2.2. I don't know where to proceed from here.
Last edited by esrever; 1 year ago
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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) = 2\sin(2x) - 2.2. I don't know where to proceed from here.
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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.
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the bear
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(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 + Ξ±) = -2.2
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esrever
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Thank you so much . But my textbook quotes an alternative method (which I found quite counter-intuitive):

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

Isn't there a easier way similar to this?
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mqb2766
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(Original post by esrever)
Thank you so much . But my textbook quotes an alternative method (which I found quite counter-intuitive):

Rewrite as it as 1=2\tan(x)-0.6\sec^2(x) and then use \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.
Last edited by mqb2766; 1 year ago
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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
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the bear
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(Original post by esrever)
Thank you so much . But my textbook quotes an alternative method (which I found quite counter-intuitive):

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