Find all values of x in [0,2pi]
All values of x:
2cos^(2) x + sinx = 1
I'd appreciate any help! Thanks!
Make everything in one term, only sin or only cos
Move everything to the right side since you want your constant number for the first term to be positive
This is a complex trinomial, so you need to use decomposition
Simplify the trinomial
Set each to zero by themselves
Since sinx = 1, you can find the x values by graph
Determine which quadrant the angles are going to be in (CAST Rule), since the value of sin is negative, it'll be in Q3 and Q4
Since there is a negative degrees (in negative coterminal), you need to find the related acute angle which is 30
Thanks for the reply! A few things, though...
In the first problem, you went from sin^2 x=1/2 to sinx=+/-1/2. Wouldn't it be +/- 1/Squareroot of 2, making the answer pi/4, not pi/6?
Also, if I have tanx=-sinx, can I divide both sides by -sinx, and just solve what's left? (-secx=0)?
Again, thanks for the help!