# a) SOLVE: cos^2 2(theta) + cos2(theta)=0 given that 0 < equal to (theta) equal to 2pieb) solve for x (0 <equal to x < equal to 2 pie): 2sin^2 x - sin x = 2 - csc x c) If sin 2x cos x +...

a) SOLVE: cos^2 2(theta) + cos2(theta)=0

given that 0 < equal to (theta) equal to 2pie

b) solve for x (0 <equal to x < equal to 2 pie):

2sin^2 x - sin x = 2 - csc x

c) If sin 2x cos x + cos 2x sin x = 1, what is the smallest possible positive value of x?

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### 1 Answer

a) You need to solve the equation `cos^2 2 theta + cos 2 theta = 0`

You need to factor out `cos 2 theta` such that:

`cos 2theta(cos 2 theta + 1) = 0`

`cos 2 theta= 0` => 2 `theta = pi/2` or 2 `theta = 3pi/2`

`theta = pi/4 or theta = 3pi/4`

**Hence, the solutions to equation `cos^2 2 theta + cos 2 theta = ` 0 in `(0,2pi)` are `theta = pi/4` or `theta = 3pi/4.` **

b) `2sin^2 x - sin x = 2 - csc x`

`2sin^2 x - sin x = 2 - 1/sin x`

`2 sin^3 x - sin^2 x - 2 sin x + 1 = 0`

You need to form two groups such that:

`(2 sin^3 x - sin^2 x) - (2 sin x- 1) = 0`

`sin^2 x(2 sin x - 1) -(2 sin x - 1) = 0`

`(2 sin x - 1)(sin^2 x - 1) = 0`

`2 sin x = 1 =gt sin x = 1/2`

`x = pi/6 or x = pi - pi/6 = 5pi/6`

`sin^2 x - 1 = 0 =gt sin x = +-1`

`sin x = 1 =gt x = pi/2`

`sin x = -1 =gt x = 3pi/2`

**Hence, the solutions to equation are x`= pi/6 , x = pi/2 , x = 5pi/6` , `x = 3pi/2.` **

`c) sin 2x cos x + cos 2x sin x = 1`

You may use the formula `sin(2x + x) = sin 2x cos x + cos 2x sin x`

Hence, `sin 3x = 1 =gt 3x = sin^(-1)(1)`

`3x = pi/2 =gt x = pi/6`

**Hence, evaluating the smallest value for x yields `x = pi/6` .**