Need Help with solving 2sinx = 1 - 2cosx

[mrs1216]

I have a test due tomorrow and I have no clue how to do this problem...please help!

Solve 2sinx=1-2cosx in the interval [0, 360)

Any help is appreciated![/list]

 

[tkhunny]

The first one is easy. Substitute cos(x) = sqrt(1-sin²(x)) and solve the resulting quadratic equation. Don't forget to check your results.

The second is a bit more challenging. Let's see what you get.

 

[mrs1216]

So substitute it so it's 2sinx=1-2(sqrt(1-sinsquaredx))???? hmm...

 

[tkhunny]

...and then...

2*sin(x) - 1 = -sqrt(1-sin²(x))

4*sin²(x) - 4*sin(x) + 1 = 1 - sin²(x)

Are we getting anywhere?

 

[mrs1216]

I'm pretty sure I got it now...thanks!

 

[tkhunny]

Did you find the second solution?

 

[soroban]

Hello, mrs1216!

Solve: \(\displaystyle \, 2\sin x\;=\;1\,-\,2\cos x\) in the interval \(\displaystyle [0^o,\,360^o)\)

A variation on TKHunny's approach . . .

We have: \(\displaystyle \,\sin x\,+\,\cos x\;=\;\frac{1}{2}\)

Square both sides: \(\displaystyle \,\sin^2x\,+\,2\sin x\cos x \,+\,\cos^2x\;=\;\frac{1}{4}\)

\(\displaystyle \;\;\)which simplifies to: \(\displaystyle \,\sin2x\,+\,1\:=\:\frac{1}{4}\;\;\Rightarrow\;\;\sin2x\:=\:-\frac{3}{4}\)

Hence, we have: \(\displaystyle \,x\;=\;\frac{1}{2}\arcsin\left(-\frac{3}{4}\right)\;\approx\;204.3^o,\;335.7^o\)


We must check our answers; squaring can produce extrneous roots.

We find that \(\displaystyle \,x\:=\;204.3^o\) is extraneous

\(\displaystyle \;\;\)and that \(\displaystyle \,x\:=\:335.7^o\) is a solution.