Question

Solve sin(t)=cos(2t) for $-4\pi \le t\le -2\pi$ .
(A). \left\\{ \dfrac{-\pi }{6},\dfrac{5\pi }{6},\dfrac{-\pi }{2} \right\\}
(B). \left\\{ \dfrac{-\pi }{3},\dfrac{-5\pi }{3},\dfrac{-\pi }{2} \right\\}
(C). \left\\{ \dfrac{-23\pi }{6},\dfrac{-19\pi }{6},\dfrac{-5\pi }{2} \right\\}
(D). \left\\{ \dfrac{-23\pi }{3},\dfrac{-19\pi }{3},\dfrac{-5\pi }{2} \right\\}
(E). \left\\{ \dfrac{-11\pi }{3},\dfrac{-7\pi }{3},\dfrac{-5\pi }{2} \right\\}

Answer 1

- Hint: In this question, we can us the relation between cos(2t) and sin(t) for solving this question and it is given as follows
$\Rightarrow \cos (2t)=1-2{{\sin }^{2}}(t)$
One more important thing is that for solving a trigonometric equation, we need to write as follows
Let us say
sinx=y
Then, for solving this equation, we will write as follows

& x=n\pi +{{(-1)}^{n}}{{\sin }^{-1}}y \\\ & \left( n\in I\ and\ {{\sin }^{-1}}y\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] \right) \\\ \end{aligned}$$ Also, for solving a quadratic equation, it is important to know the following formula which is called the quadratic formula $$roots=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ for a quadratic equation $$a{{x}^{2}}+bx+c=0$$. Firstly, we will convert the given expression in terms of sin function and then we will use the above mentioned formula to solve and get the value from the equation. _Complete step-by-step solution_ - As mentioned in the question, we have to solve the given equation and hence, we have to get the values of t. Now, as mentioned in the hint, we will first convert the given equation in terms of sin function as follows $$\begin{aligned} & \Rightarrow \sin t=\cos 2t \\\ & (\cos (2t)=1-2{{\sin }^{2}}(t)) \\\ & \Rightarrow \sin t=1-2{{\sin }^{2}}(t) \\\ & \Rightarrow 2{{\sin }^{2}}(t)+\sin t-1=0 \\\ \end{aligned}$$ Now, let sin(t) be equal to A and so, we will solve the quadratic equation in A as follows using the quadratic formula which is $$roots=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ for a quadratic equation $$a{{x}^{2}}+bx+c=0$$. $$\begin{aligned} & \Rightarrow 2{{A}^{2}}+A-1=0 \\\ & \Rightarrow A=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times 2\times \left( -1 \right)}}{2\times 2} \\\ & \Rightarrow A=\dfrac{-1\pm \sqrt{1+8}}{2\times 2} \\\ & \Rightarrow A=\dfrac{-1\pm \sqrt{9}}{2\times 2} \\\ & \Rightarrow A=\dfrac{-1\pm 3}{4} \\\ & \Rightarrow A=\dfrac{-1-3}{4},\dfrac{-1+3}{4} \\\ & \Rightarrow A=\dfrac{-4}{4},\dfrac{2}{4} \\\ & \Rightarrow A=-1,\dfrac{1}{2} \\\ \end{aligned}$$ Now, we have got two values of A that means we have got two values of sin(t), so, we will form two cases as follows Case 1:-sin(t)=1 Now, using the method given in the hint, we can solve this equation as follows $$\begin{aligned} & \Rightarrow t=n\pi +{{(-1)}^{n}}{{\sin }^{-1}}(-1) \\\ & \left( n\in I\ and\ {{\sin }^{-1}}(-1)\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] \right) \\\ & \Rightarrow t=n\pi +{{(-1)}^{n}}\left( \dfrac{-\pi }{2} \right) \\\ & \left( n\in I\ and\ \dfrac{-\pi }{2}\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] \right) \\\ & \Rightarrow t=\dfrac{-\pi }{2},\dfrac{-3\pi }{2},\dfrac{-5\pi }{2} \\\ \end{aligned}$$ Case 2:- $$sin\left( t \right)=\dfrac{1}{2}$$ Now, using the method given in the hint, we can solve this equation as follows $$\begin{aligned} & \Rightarrow t=n\pi +{{(-1)}^{n}}{{\sin }^{-1}}\dfrac{1}{2} \\\ & \left( n\in I\ and\ {{\sin }^{-1}}\dfrac{1}{2}\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] \right) \\\ & \Rightarrow t=n\pi +{{(-1)}^{n}}\left( \dfrac{\pi }{6} \right) \\\ & \left( n\in I\ and\ \left( \dfrac{\pi }{6} \right)\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] \right) \\\ & \Rightarrow t=\dfrac{\pi }{6},\dfrac{5\pi }{6},\dfrac{-23\pi }{6},\dfrac{-19\pi }{6} \\\ \end{aligned}$$ Now, we can see that only option (B) has the correct solutions of t amongst all the options. Hence, the correct answer is option (B). Note:-In these questions in which trigonometric equations are to be solved, it is very important to know the standard formulae for sin, cos and tan functions as they can come in handy in such situations. Also, the students can commit calculation errors, so, one should be extra careful while solving the question and while doing the calculations.