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Question: How do you solve \(\sin 2x - 1 = 0\) from \(\left[ {0,\;360} \right]\) ?...

How do you solve sin⁡2x−1=0\sin 2x - 1 = 0 from [0,  360]\left[ {0,\;360} \right] ?

Explanation

Solution

First simplify the given trigonometric equation and then find the value in the interval given interval and there are two chances of being the interval in degrees or in radians which is not mentioned in the problem so, consider both the units and solve accordingly.

Complete step by step solution:
In order to solve the given trigonometric equation sin⁡2x−1=0\sin 2x - 1 = 0 in the interval [0,  360]\left[ {0,\;360} \right], we will first simplify the given trigonometric equation (that is all variables one side and the constants on the another). Now from the given trigonometric equation
⇒sin⁡2x−1=0 ⇒sin⁡2x=1 \Rightarrow \sin 2x - 1 = 0 \\\ \Rightarrow \sin 2x = 1 \\\
Taking inverse function of sine both sides we will get,
⇒sin⁡−1(sin⁡2x)=sin⁡−1(1) ⇒2x=sin⁡−1(1) \Rightarrow {\sin ^{ - 1}}\left( {\sin 2x} \right) = {\sin ^{ - 1}}(1) \\\ \Rightarrow 2x = {\sin ^{ - 1}}(1) \\\
Now from the inverse trigonometric table, we know that the value of sine inverse of one is equals to 900  or  π2{90^0}\;or\;\dfrac{\pi }{2} in the principle interval [0,  3600]  or  [0,  2π]\left[ {0,\;{{360}^0}} \right]\;or\;\left[ {0,\;2\pi } \right], but here in this question we are not sure if the interval [0,  360]\left[ {0,\;360} \right] is given in degrees or radians, so we will take solve for both one by one.
Solving the equation for case I: That is taking the interval [0,  360]\left[ {0,\;360} \right] in degrees,
⇒[0,  3600]\Rightarrow \left[ {0,\;{{360}^0}} \right]
⇒2x=sin⁡−1(1)\Rightarrow 2x = {\sin ^{ - 1}}(1)
In the interval [0,  3600]\left[ {0,\;{{360}^0}} \right] there is only one solution for sin⁡−1(1){\sin ^{ - 1}}(1)

\Rightarrow x = \dfrac{{{{90}^0}}}{2} \\\ \Rightarrow x = {45^0} \\\ $$ That is we get the solution $$x = {45^0}$$ for this case. Solving the equation for case II: That is taking the interval $\left[ {0,\;360} \right]$ in radians, $$ \Rightarrow 2x = {\sin ^{ - 1}}(1)$$ Now in the interval $\left[ {0,\;360} \right]$ there will be many solutions for $${\sin ^{ - 1}}(1)$$ which can be given as $$ \Rightarrow 2x = 2n\pi + \dfrac{\pi }{2},\;where\;n \in W\;{\text{and}}\;2n\pi + \dfrac{\pi }{2} < 360$$ Finding value for $n$ $$2n\pi < 360 - \dfrac{\pi }{2} \\\ 2n\pi < 358.43 \\\ n < \dfrac{{358.43}}{{2\pi }} \\\ n < 57.046 \\\ $$ So $n < 57.056\;{\text{and}}\;n \in W\;\therefore n = 57$ $\therefore x = 2n\pi + \dfrac{\pi }{2},\;{\text{where}}\;n \in W\;{\text{and}}\;n \leqslant 57$ is solution for second case. **Note:** Generally when only the digits are given in the arguments of angle then we consider the radian unit for them, but we are familiar that whenever radian unit is being used in questions the digits are associated with pi because general values of trigonometric function are given with pi-digit radians. That’s why we have considered both the cases.