# Trig Equation

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#1
0
1 year ago
#2
double angle formula?
0
1 year ago
#3
cos^2(x)-2cos(x)sin(x)+0.6=0
sin^2(x)=0.6
1
#4
(Original post by mqb2766)
double angle formula?
Do you mean rewriting 2sin(x)cos(x) as sin(2x)?
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#5
(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?
0
1 year ago
#6
(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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#7
(Original post by mqb2766)
yes and similar for cos^2(x)
I tried it but got . I don't know where to proceed from here.
Last edited by esrever; 1 year ago
0
1 year ago
#8
Good point :-). Give me a sec.
0
1 year ago
#9
(Original post by mqb2766)
Good point :-). Give me a sec.
Convert the cos and sin into a single
Rsin(2x + alpha)
then solve.
1
1 year ago
#10
(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
1
#11
Thank you so much . But my textbook quotes an alternative method (which I found quite counter-intuitive):

Rewrite as it as and then use .

Isn't there a easier way similar to this?
0
1 year ago
#12
(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 and then use .

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
0
#13
(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 0
1 year ago
#14
(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 and then use .

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.
0
#15
(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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