Given that cos(θ + (π/3))= (1/2) find the values of θ inthe range 0°≤ θ≤360°


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Question Number 41444 by Rio Michael last updated on 07/Aug/18

$G i v e n t h a t c o s (θ + \frac{π}{3}) = \frac{1}{2} f i n d t h e v a l u e s o f θ i n t h e r a n g e 0 ° ⩽ θ ⩽ 360 °$
Commented by maxmathsup by imad last updated on 07/Aug/18

$c o s (θ + \frac{π}{3}) = \frac{1}{2} \Leftrightarrow c o s (θ + \frac{π}{3}) = c o s (\frac{π}{3}) \Leftrightarrow θ + \frac{π}{3} = \frac{π}{3} + 2 k π o r$ $θ + \frac{π}{3} = - \frac{π}{3} + 2 k π (k \in Z) \Leftrightarrow θ = 2 k π o r θ = - \frac{2 π}{3} + 2 k π n o w l e t s o v e i n [0, 2 π]$ $0 ⩽ 2 k π ⩽ 2 π \Rightarrow k = 0 o r k = 1$ $0 ⩽ - \frac{2 π}{3} + 2 k π ⩽ 2 π \Rightarrow 0 ⩽ - \frac{2}{3} + 2 k ⩽ 2 \Rightarrow \frac{2}{3} ⩽ 2 k ⩽ 2 + \frac{2}{3} \Rightarrow$ $\frac{1}{3} ⩽ k ⩽ \frac{4}{3} \Rightarrow 0, . . . ⩽ k ⩽ 1, . . \Rightarrow k = 1 \Rightarrow x = 2 π - \frac{2 π}{3} = \frac{4 π}{3} t h e s l u t i o n s a r e$ $0, 2 π a n d \frac{4 π}{3}$
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