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Question: How do you solve the equation on the interval \( \left[ {0,\;2\pi } \right) \) for \( 2\sin t\cos t ...

How do you solve the equation on the interval [0,  2π)\left[ {0,\;2\pi } \right) for 2sin⁡tcos⁡t+sin⁡t−2cos⁡t−1=0?2\sin t\cos t + \sin t - 2\cos t - 1 = 0?

Explanation

Solution

Hint : To solve the given trigonometric equation in the given interval, first of all simplify the equation such that similar terms are being held together in order to factorize the right hand side expression then equate factorized terms with zero separately and solve for different cases and then take the union of all solutions to get the final solution.

Complete step-by-step answer :
In order to find the solution for the given trigonometric equation 2sin⁡tcos⁡t+sin⁡t−2cos⁡t−1=02\sin t\cos t + \sin t - 2\cos t - 1 = 0 , we will factorize the right hand side expression by grouping similar terms as follows
⇒2sin⁡tcos⁡t+sin⁡t−2cos⁡t−1=0 ⇒sin⁡t(2cos⁡t+1)−1(2cos⁡t+1)=0 ⇒(2cos⁡t+1)(sin⁡t−1)=0   \Rightarrow 2\sin t\cos t + \sin t - 2\cos t - 1 = 0 \\\ \Rightarrow \sin t\left( {2\cos t + 1} \right) - 1\left( {2\cos t + 1} \right) = 0 \\\ \Rightarrow \left( {2\cos t + 1} \right)\left( {\sin t - 1} \right) = 0 \;
So here we can see that, we are getting two factors of the given equation, so equating both factors separately with zero in order to get the solution,
Comparing (2cos⁡t+1)\left( {2\cos t + 1} \right) with zero, we will get
⇒2cos⁡t+1=0 ⇒cos⁡t=−12   \Rightarrow 2\cos t + 1 = 0 \\\ \Rightarrow \cos t = - \dfrac{1}{2} \;
Now, we know the general solution of cos⁡t=−12\cos t = - \dfrac{1}{2} is given as
t=2nπ±2π3,  where  n∈It = 2n\pi \pm \dfrac{{2\pi }}{3},\;{\text{where}}\;n \in I
But in the given interval i.e. [0,  2π)\left[ {0,\;2\pi } \right) the value of n=0  and  n=1n = 0\;{\text{and}}\;n = 1 we will get
t=2π3  and  4π3t = \dfrac{{2\pi }}{3}\;{\text{and}}\;\dfrac{{4\pi }}{3}
Now, comparing (sin⁡t−1)\left( {\sin t - 1} \right) with zero, we will get
⇒sin⁡t−1=0 ⇒sin⁡t=1   \Rightarrow \sin t - 1 = 0 \\\ \Rightarrow \sin t = 1 \;
Now, we know the general solution of sin⁡t=1\sin t = 1 is given as
t=2nπ±π2,  where  n∈It = 2n\pi \pm \dfrac{\pi }{2},\;{\text{where}}\;n \in I
But in the given interval i.e. [0,  2π)\left[ {0,\;2\pi } \right) the value of n=0n = 0 we will get
t=π2t = \dfrac{\pi }{2}
Therefore the final solution will be given as the union of both the solutions that is

t=π2,  2π3  and  4π3t = \dfrac{\pi }{2},\;\dfrac{{2\pi }}{3}\;{\text{and}}\;\dfrac{{4\pi }}{3} is the required solution.

Note : To get the required value of the angle, the trigonometric table should be remembered, the angle can be written as degrees or radians. In what quadrants to get the sign of the angle, it must be remembered which trigonometric functions are positive and negative.