Solve for x in the range 0 ≤ x ≤2π the equations a) cos(x + (π/3)) = 0 b) sin x = cos x. c) sin 2x + 2sin x = 1 + cos x


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Question Number 39587 by Rio Mike last updated on 08/Jul/18

$S o l v e f o r x i n t h e r a n g e 0 ⩽ x ⩽ 2 π$ $t h e e q u a t i o n s$ $a) c o s (x + \frac{π}{3}) = 0$ $b) s i n x = c o s x .$ $c) s i n 2 x + 2 s i n x = 1 + c o s x$
	Answered by Joel579 last updated on 08/Jul/18

	$\cos (x + \frac{π}{3}) = 0$ $x + \frac{π}{3} = n π + \frac{π}{2} (n = 0, 1)$ $x = \frac{π}{6} \lor x = \frac{7 π}{6}$
	Answered by Joel579 last updated on 08/Jul/18

	$\sin x = \cos x$ $\tan x = 1$ $x = n π + \frac{π}{4} (n = 0)$ $x = \frac{π}{4}$
	Answered by Joel579 last updated on 08/Jul/18

	$2 \sin x \cos x + 2 \sin x = 1 + \cos x$ $2 \sin x (\cos x + 1) = 1 + \cos x$ $2 \sin x (1 + \cos x) - (1 + \cos x) = 0$ $(2 \sin x - 1) (1 + \cos x) = 0$ $\sin x = \frac{1}{2} \to x = \frac{π}{6}, \frac{5 π}{6}$ $\cos x = - 1 \to x = π$
	Answered by MJS last updated on 08/Jul/18
	$(a) x+(π/3)=arccos 0 ={(π/2); ((3π)/2)} x={(π/2)−(π/3); ((3π)/2)−(π/3)}={(π/6); ((7π)/6)} (b) think of the circle x^2 +y^2 =1 again cos θ =sin θ means we′re looking for the points (((cos θ)),((sin θ)) ) = ((p),(q) ) with p=q ⇒ ⇒ p=q=((√2)/2) ∨ p=q=−((√2)/2) ⇒ ⇒ θ={−((5π)/4); (π/4)} (c) sin 2x=2sin x cos x 2sc+2s=1+c 2sc−c+2s−1=0 c(2s−1)+2s−1=0 (2s−1)(c+1)=0 2sin x −1=0 ∨ cos x +1=0 sin x =(1/2) ∨ cos x =−1 x=({(π/6); ((5π)/6)} ∪ {π})={(π/6); ((5π)/6); π}$
	$(a) x + \frac{π}{3} = \arccos 0 = {\frac{π}{2}; \frac{3 π}{2}}$ $x = {\frac{π}{2} - \frac{π}{3}; \frac{3 π}{2} - \frac{π}{3}} = {\frac{π}{6}; \frac{7 π}{6}}$ $(b) think of the circle x^{2} + y^{2} = 1 again$ $\cos θ = \sin θ means {we}^{'} re looking for the$ $points (\begin{matrix} \cos θ \\ \sin θ \end{matrix}) = (\begin{matrix} p \\ q \end{matrix}) with p = q \Rightarrow$ $\Rightarrow p = q = \frac{\sqrt{2}}{2} \lor p = q = - \frac{\sqrt{2}}{2} \Rightarrow$ $\Rightarrow θ = {- \frac{5 π}{4}; \frac{π}{4}}$ $(c) \sin 2 x = 2 \sin x \cos x$ $2 s c + 2 s = 1 + c$ $2 s c - c + 2 s - 1 = 0$ $c (2 s - 1) + 2 s - 1 = 0$ $(2 s - 1) (c + 1) = 0$ $2 \sin x - 1 = 0 \lor \cos x + 1 = 0$ $\sin x = \frac{1}{2} \lor \cos x = - 1$ $x = ({\frac{π}{6}; \frac{5 π}{6}} \cup {π}) = {\frac{π}{6}; \frac{5 π}{6}; π}$
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