# Thread: Solve: cos^−1(2x) + sin^−1(x) = π/3

1. Originally Posted by pka
\begin{align*} \arccos (2x) + \arcsin (x) &= \frac{\pi }{3}\\ \cos (\arccos (2x) + \arcsin (x)) &= \cos \left( {\frac{\pi }{3}} \right)\\ \cos (\arccos (2x))\cos (\arcsin (x)) - \sin (\arccos (2x))\sin (\arcsin (x)) &= \frac{1}{2} \end{align*}

Now you should know these; if not then you need sit-down help.
$\cos (\arccos (2x))=2x$ & $\cos (\arcsin (x))=\sqrt{1-x^2}$

If you do understand those then show us by posting the solution.

I get:

2x*√(1-x^2) - x/(√1-4x^2) = 1/2

I tried simplifying further by multiplying by 2 and by (√1-4x^2) but, I can't see a solution for x.
I used a graphics calculator to draw a graph of y = 2x*√(1-x^2) - x/(√1-4x^2) and then used the trace function to trace the graph to y=0.5 to find x. I got x = -0.47
How can I solve for x using just pen & paper?

2. Originally Posted by Bobby Bones
I get:

2x*√(1-x^2) - x/(√1-4x^2) = 1/2
It appears as if you are dividing when there is no division.

It should be $2x\sqrt{1-x^2}-x\sqrt{1-4x^2}=\dfrac{1}{2}$ But that is a beast of an equation to solve.
SEE HERE.

3. Originally Posted by pka
It appears as if you are dividing when there is no division.

It should be $2x\sqrt{1-x^2}-x\sqrt{1-4x^2}=\dfrac{1}{2}$ But that is a beast of an equation to solve.
SEE HERE.

Thanks, you are right. I put my values for opp and hyp the wrong way round when i calculated sin(theta) from cos(theta)=2x.
So what should I answer for the original question? It is asking me to "solve" the original equation.
The question is the first/easiest question on a A-Level standard trig exam past paper and the question is worth 4 marks.

4. Beer soaked reckoning follows.
Originally Posted by pka
But that is a beast of an equation to solve.
SEE HERE.
You can say that again!
Tried it with me booze enhanced vision and got
208x^4-40x^2+1=0 after two "squaring" procedures.
Blind application of the quadratic formula gives one real answer equivalent (can't figure out how to manipulate "my" answer to line up with WA and TI89) to WA and three non real answers.
Maybe somebody here has some idea how to rearrange the real number solution to
208x^4-40x^2+1=0 so that it looks like that somewhat reduced/simplified WA answer.
Originally Posted by Bobby Bones
Thanks, you are right. I put my values for opp and hyp the wrong way round when i calculated sin(theta) from cos(theta)=2x.
So what should I answer for the original question? It is asking me to "solve" the original equation.
The question is the first/easiest question on a A-Level standard trig exam past paper and the question is worth 4 marks.
Only 4 marks?
And that's supposed to be easy?

5. Beer soaked correction follows.
Originally Posted by jonah2.0
Maybe somebody here has some idea how to rearrange the real number solution to
208x^4-40x^2+1=0 so that it looks like that somewhat reduced/simplified WA answer.
After three more beer bottles, finally got my answer to line up with WAs.
As for TI89s sqr[13(2*sqr(3)+5)]/26, I am stumped.

6. Originally Posted by jonah2.0
After three more beer bottles, finally got my answer to line up with WAs.
As for TI89s sqr[13(2*sqr(3)+5)]/26, I am stumped.
u = √3

√[13(2u+5)] / 26
That TI89 expression is WA's x = √[(2u+5)/13] / 2 solution (there are 3 others).

In other words:
√[13(2u+5)] / 26 = √[(2u+5)/13] / 2 = .4034491106775367282.......whoa!!

See ya Sir Jonah: I need my 4th coffee...

7. Originally Posted by Denis
u = √3

√[13(2u+5)] / 26
That TI89 expression is WA's x = √[(2u+5)/13] / 2 solution (there are 3 others).

In other words:
√[13(2u+5)] / 26 = √[(2u+5)/13] / 2 = .4034491106775367282.......whoa!!

See ya Sir Jonah: I need my 4th coffee...
Beautiful work. You get your 4 marks. Btw, the exam paper has 7 questions in total and a 75 minute time limit. No beer allowed 😁

8. Beer soaked complaint follows.
Originally Posted by Bobby Bones
Btw, the exam paper has 7 questions in total and a 75 minute time limit. No beer allowed 😁
WTF!!!???
Coffee allowed but not beer?
Well you can always drink a beer or two before going in the exam room.
Also, I'd love to see the rest of those 6 other questions just for the fun of it; you know, to see how much dinosaurs could tackle them modern exam questions.

9. Originally Posted by jonah2.0
Beer soaked complaint follows.

WTF!!!???
Coffee allowed but not beer?
Well you can always drink a beer or two before going in the exam room.
Also, I'd love to see the rest of those 6 other questions just for the fun of it; you know, to see how much dinosaurs could tackle them modern exam questions.
I posted a copy of one full paper, just yesterday, please check - the thread should be jusy a few places above this one.
I'll attach the full paper that contains this question here. I would be grateful if you would share the answers with your methods included. I am doing all this to help a poor, Indonesian student. I am happy to help where i can. Just ask.
I've attached all the 4 past papers and a copy of the formula sheet they are given:

10. can u read the attachments? They are too blurred when i click on them.

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