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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 + \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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