+14 1) Solve the following equation, where theta is in the reals.
4cos(2theta)+3=0
2)Solve for x over the interval of [0, 2pi]
-2 tan^2x=0
Please include a little bit of explanation, thanks
+128473 4cos(2 theta)+3=0 subtract 3 from both sides
4 cos (2 theta) = - 3 divide both sides by 4
cos (2 theta) = -3 / 4
cos (theta) = -3/4 at ≈ [ 2.42 ± 2pi ] rads and at ≈ [3.86 ± 2pi ] rads
So
cos ( 2 theta) = ( [ 2.42 ± 2pi ] / 2 ) rads and at ( [ 3.86 ± 2pi ] / 2 rads ) =
[ 1.21 ± pi ] rads and [ 1.93 ± pi ] rads
+128473 2)Solve for x over the interval of [0, 2pi]
-2 tan^2x=0
Divide both sides by - 2
tan^2 x = 0 take the square root and we get that
tan x = 0
This happens at x = 0 , pi and 2pi rads on the requested interval
+128473 Comet's second queston is actually this :
2)Solve for x over the interval of [0, 2pi]
-2 + tan^2x=0
Add 2 to both sides
tan^2x = 2 taske both square roots
tan x = ± √2
We either have that
tan x = √2
Take the tan inverse to find the angle in rads
arcctan (√2) ≈ .955 rads this will also occur at pi + .955 rads = 4.096 rads
Also
tan x = -√2
arctan ( -√2) ≈ -.955 rads = [2pi - .955] rads ≈ 5.328 rads
And once more at [5.328 - pi] rads ≈ 2.19 rads
So....the four answers are ≈ [ .955 , 2.19, 4.096 and 5.328 ] rads
