![\cos(2x) + \dfrac{\sin x}{2} = \cos^2 x](/tex.php?f=\cos(2x) + \dfrac{\sin x}{2} = \cos^2 x)
for 0°≤x≤360°answer and explanationthe answers are x = 0°, 30°, 150°, 180°, 360°this can be solved using trigonometric identities.notice that we have a first-power sine term.the terms cos(2x) and cos²x all have some sort of identity that can change them into something with sine. therefore, it will be useful to have everything as sine.the pythagorean identity links cos²x and sin²x via the identity sin²x + cos²x = 1 ⇒ cos²x = 1 - sin²xthat allows us to convert cosine square into sine squareda cosine double identity to use: cos(2x) = 1 - 2sin²xnow we can convert everything in terms of sine:
![\begin{aligned} \cos(2x) + \dfrac{\sin x}{2} & = \cos^2 x \\ 1 - 2\sin^2 x + \tfrac{1}{2}\sin x & = 1 - \sin^2 x \end{aligned}](/tex.php?f=\begin{aligned} \cos(2x) + \dfrac{\sin x}{2} & = \cos^2 x \\ 1 - 2\sin^2 x + \tfrac{1}{2}\sin x & = 1 - \sin^2 x \end{aligned})
subtracting 1 from both sides results in
![- 2\sin^2 x + \tfrac{1}{2}\sin x = - \sin^2 x](/tex.php?f=- 2\sin^2 x + \tfrac{1}{2}\sin x = - \sin^2 x )
get rid of the 2sin²x term on the leftadding 2sin²x to both sides results in
![\tfrac{1}{2}\sin x = \sin^2 x](/tex.php?f=\tfrac{1}{2}\sin x = \sin^2 x)
subtracting (1/2)sin x from both sides results in
![0 = \sin^2 x - \tfrac{1}{2}\sin x](/tex.php?f=0 = \sin^2 x - \tfrac{1}{2}\sin x)
factor out sin x
![0 = \sin x \left(\sin x - \tfrac{1}{2}\right)](/tex.php?f=0 = \sin x \left(\sin x - \tfrac{1}{2}\right) )
we have one side of the equation fully factored and the other side equal to zero.therefore, either sin x = 0 or sin x - 1/2 = 0 because zero times anything is zero (zero factor property). solving the first one sin x = 0in a unit circle, sine is defined as the ratio of the y-coordinate of the point of intersection of the terminal arm to the radiusby the unit circle, the solutions, for 0°≤ x ≤ 360°, include 0°, 180°, and 360° (y-coordinate is 0 for those angles)solving the other equation: sin x - 1/2 = 0move and you get sin x = 1/2. by the unit circle, the solutions include 30° and 150°thus the final solution is x = 0°, 30°, 150°, 180°, 360°you can test all this in the original equation and all of them will work