5cos^2 x + sin^2 x = 4 , 0<x<2pi

[JAsh]

i've tried doing it but i have no clue!

please help!!

 

[galactus]

$\displaystyle 5cos^{2}(x)+sin^{2}(x)=4$

Sub in $\displaystyle cos^{2}(x)=1-sin^{2}(x)$

$\displaystyle 5(1-sin^{2}(x))+sin^{2}(x)=4$

$\displaystyle 5-5sin^{2}(x)+sin^{2}(x)=4$

Can you finish?.

You will have several solutions in your interval.

 

[Mrspi]

i've tried doing it but i have no clue!

please help!!

From the Pythagorean Identity,

sin[sup:3tp6glac]2[/sup:3tp6glac] x + cos[sup:3tp6glac]2[/sup:3tp6glac] x = 1

you can get this:

cos[sup:3tp6glac]2[/sup:3tp6glac] x = 1 - sin[sup:3tp6glac]2[/sup:3tp6glac] x

Substitute (1 - sin[sup:3tp6glac]2[/sup:3tp6glac] x) for cos[sup:3tp6glac]2[/sup:3tp6glac] x

Then you'll have an equation involving just ONE function.

 

[JAsh]

$\displaystyle 5cos^{2}(x)+sin^{2}(x)=4$

Sub in $\displaystyle cos^{2}(x)=1-sin^{2}(x)$

$\displaystyle 5(1-sin^{2}(x))+sin^{2}(x)=4$

$\displaystyle 5-5sin^{2}(x)+sin^{2}(x)=4$

Can you finish?.

You will have several solutions in your interval.

i have no idea how to carry on!
i dont know how ot get the solutions

 

[galactus]

Continuing where I left off:

$\displaystyle 5-5sin^{2}(x)+sin^{2}(x)=4$

You can certainly see to add the $\displaystyle sin^{2}(x)$'s?.

$\displaystyle 5-4sin^{2}(x)=4$

$\displaystyle -4sin^{2}(x)=-1$

$\displaystyle sin^{2}(x)=\frac{1}{4}$

Now, solve for x in the interval $\displaystyle [0,2\pi]$. I am sorry to say, if you are clueless at this point you really should see your instructor.