Solve Eqtn for 0º ≤ θ ≤ 360º: sin^2(x) + 2cos^2(x) = 2

[ninatemple]

Solve Eqtn for 0º ≤ θ ≤ 360º: sin^2(x) + 2cos^2(x) = 2

1. sin^2(x) + 2cos^2(x) = 2

 

[stapel]

1. sin^2(x) + 2cos^2(x) = 2

What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

For instance, in the usual manner, you noted that the left-hand side (LHS) is more complicated, so you started there. You noted that everything was already stated in terms of sines and cosines, so there was no conversion necessary. Then you noted that either of the squared terms could be stated in a form using only the other trig function, so you made a substitution, and... then what?

Please be complete. Thank you!

 

[Mrspi]

1. sin^2(x) + 2cos^2(x) = 2

Since it has been more than a day since you posted this question, I'll give you a hint.

Would you agree that the left side could be written like this?

sin² x + cos² x + cos² x = 2

What could you use to replace sin² x + cos² x ?

Could you then solve the resulting equation for x in the specified interval?

 

[ninatemple]

Since it has been more than a day since you posted this question, I'll give you a hint.

Would you agree that the left side could be written like this?

sin² x + cos² x + cos² x = 2

What could you use to replace sin² x + cos² x ?

Could you then solve the resulting equation for x in the specified interval?

Thank you, I got it now.