Question

Solve in interval of [0,2pi)

2cosx-3tanx=0

Answer 1

I think you can convert \(\tan x\) into \(\dfrac{\sin x}{\cos x}\), then put everything on the same denominator of \(\cos x\).

Then you should be able to convert \(\cos^2 x\) by using \(\cos^2x + \sin^2x=1\) , so that your whole function is in \(\sin x\)'s , and then solve the equation accordingly.

Answer 2

I can provide more guidance if you need it. I guess my help is a bit "vague"

Answer 3

Yes please. When I put everything under one denominator I have (2cos^2x-3sinx)/cosx. I don't know where to go from there.

Answer 4

Yeah that's good. Try converting \(\cos^2 x\) by using \(1-\sin^2x\) , which comes from \(\cos^2x+\sin^2 = 1\)

Answer 5

\(\sin^2x\) **

Answer 6

I have (2-sin^2x-3sinx)/cosx, which arranged it would be (-sin^2x-3sinx+2)/cosx

Answer 7

I think you forgot a coefficient of 2 infront of \(\sin^2x\)

Answer 8

Right.

Answer 9

But once you do that, notice that:
\[\frac{-2\sin^2x - 3\sin x +2}{\cos x} =0 \] implies
\( -2\sin^2x - 3\sin x +2=0\), And then you say let \(\sin x = y\), and realize that you have a quadratic function that you can solve for \(y\)

Answer 10

Yes, but what about the cosx?

Answer 11

You can think that you're cross-multiplying your fractions. Like you can say the right side is \(\dfrac{0}{1}\), and you cross multiply so that you get \(1(-2\sin^2 x - 3 \sin x + 2)\) = \(0 *\cos(x)\) , so cos(x) disappears

Answer 12

Ah, thanks.

Answer 13

so the solutions I have are pi/6 and 5pi/6

Answer 14

Yeah I get that as well :)

Answer 15

Awesome, thank you!

Answer 16

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