Question 1202973: Hi,
Can you help me with this one?
Use the Auxiliary Angle Method to solve 3cos(2x) - 2sin(2x) =1 on 0≤x≤2π, to 3DP.
I get four solutions, is this correct?
Thank you
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52855)   (Show Source):
You can put this solution on YOUR website! .
You can easily check it graphically / visually on your own.
Go to web-site www.desmos.com/calculator.
Print there the expression of the left side - you will see the plot of the left side function.
In the next line, print expression for the right side, which is a constant "1".
Look at the interval [0,2pi) and count how many intersection points do you see there.
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Comment from student : Thank you for your help. I did check through DESMOS and think I am correct.
My response : I am glad that you mastered DESMOS plotting tool and learned this method
of checking your solutions.
Come again to this forum soon to learn something new ( ! )
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Another approach would be to plot the function
f(x) = 3cos(2x) - 2sin(2x) - 1
and look for the x intercepts aka roots.
I have done so with this Desmos plot
https://www.desmos.com/calculator/zeflra7phh
The blue shaded region represents the interval  aka [0, 2pi]
2pi = 6.28 approximately
Each x intercept is labeled on the graph, and also in the left-side panel.
With Desmos, you can click on the roots to have the coordinates show up.
The four approximate solutions in that interval are:
0.351
2.203
3.492
5.344
To confirm each solution, plug them back one at a time into f(x)
For instance, let's try x = 0.351
f(x) = 3cos(2x) - 2sin(2x) - 1
f(0.351) = 3cos(2*0.351) - 2sin(2*0.351) - 1
f(0.351) = -0.0008354950597464
There's some slight rounding error going on. But it's close enough to 0.
Make sure your calculator is in radian mode.
I'll let you check the others.
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