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5cos^2 x + sin^2 x = 4 , 0<x<2pi
- Thread starter JAsh
- Start date
\(\displaystyle 5cos^{2}(x)+sin^{2}(x)=4\)
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.
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.
JAsh said: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.
galactus said:\(\displaystyle 5cos^{2}(x)+sin^{2}(x)=4\)
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
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.
\(\displaystyle 5-5sin^{2}(x)+sin^{2}(x)=4\)
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.