Question

How to solve $\cos x-2\sin x=1$ for $0\le x\le 2\pi $ ?

Answer 1

Hint: We see that, in this question, we have to find the value of x , where a trigonometric equation is given to us with the value of x lies in some interval. So, we start solving this problem by adding 2sinx and then we square both sides of the equation and make some necessary calculations. After that, we subtract ${{\cos }^{2}}x$ on both sides of the equation and then apply trigonometric identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$. After further simplifications, we let $\sin x=t$ and then solve the equation for t, after getting values of t, we again substitute those values in $\sin x=t$ and solve for x under the constraint of interval given in the question, which is the required answer to our problem.

Complete answer:
According to the question, we have to solve the trigonometric equation, to get the value of x.
So, we use the trigonometric identities and formula, to get the required solution.
The equation given is $\cos x-2\sin x=1$ ---------- (1)
Firstly, we add $2\sin x$ on both sides in the equation (1), we get
$\Rightarrow \cos x-2\sin x+2\sin x=1+2\sin x$
As we know, the same terms of opposite signs cancel out, therefore we get
$\Rightarrow \cos x=1+2\sin x$
Now, we will square both the sides in the above equation, to get
$\Rightarrow {{(\cos x)}^{2}}={{(1+2\sin x)}^{2}}$
Now, we will apply the algebraic identity ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ on the RHS of the above equation, we get
$\begin{align}
  & \Rightarrow {{\cos }^{2}}x={{1}^{2}}+2.(1).(2\sin x)+{{(2\sin x)}^{2}} \\
 & \Rightarrow {{\cos }^{2}}x=1+4\sin x+4{{\sin }^{2}}x \\
\end{align}$
Now, subtract both sides of the equation by ${{\cos }^{2}}x$ , we get
$\Rightarrow {{\cos }^{2}}x-{{\cos }^{2}}x=1+4\sin x+4{{\sin }^{2}}x-{{\cos }^{2}}x$
As we know, the same terms with opposite signs cancel out, therefore, we get
$\Rightarrow 0=1+4\sin x+4{{\sin }^{2}}x-{{\cos }^{2}}x$
We will use the trigonometric identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ in the above equation, we get
$\Rightarrow 0=1+4\sin x+4{{\sin }^{2}}x-(1-{{\sin }^{2}}x)$
On further simplification, we get
$\Rightarrow 0=1+4\sin x+4{{\sin }^{2}}x-1+{{\sin }^{2}}x$
Now, again same terms with opposite signs cancel out and make necessary calculations, therefore we get
$\Rightarrow 0=4\sin x+5{{\sin }^{2}}x$ --------- (2)
Now, let us suppose that $\sin x=t$ ----- (3)
Thus, we will substitute this value in equation (2), we get
$\Rightarrow 0=4t+5{{t}^{2}}$
Taking factors on the right-hand side of the above equation, to get
$\Rightarrow 0=t(4+5t)$
$\Rightarrow t=0$ ---- (4) or $\Rightarrow 4+5t=0$ -------- (5)
So, let us solve equation (5) by subtracting 4 on both sides of the equation and then cancel out the same terms with opposite signs and after that, we divide 5 on both LHS and RHS of the equation, we get
$\begin{align}
  & \Rightarrow 4+5t-4=0-4 \\
 & \Rightarrow 5t=-4 \\
 & \Rightarrow \dfrac{5}{5}t=-\dfrac{4}{5} \\
\end{align}$
Therefore, we get
$\Rightarrow t=-\dfrac{4}{5}$ ----------- (6)
Now, from equation (4) and (6), we get the value of t as $t=0$ or $t=-\dfrac{4}{5}$
Thus, we now substitute equation (4) in equation (3), we get
$\begin{align}
  & \Rightarrow \sin x=t \\
 & \Rightarrow \sin x=0 \\
 & \Rightarrow x={{\sin }^{-1}}(0) \\
\end{align}$
As the range of x given in the question is $0\le x\le 2\pi $, therefore we get
$\Rightarrow x=0$ or $\Rightarrow x=\pi $ or $\Rightarrow x=2\pi $
Similarly, we substitute equation (6) in equation (3), we get
$\begin{align}
  & \Rightarrow \sin x=t \\
 & \Rightarrow \sin x=-\dfrac{4}{5} \\
 & \Rightarrow x={{\sin }^{-1}}\left( -\dfrac{4}{5} \right) \\
\end{align}$
As the range of x given in the question is $0\le x\le 2\pi $, therefore we get
$\Rightarrow x={{\sin }^{-1}}\left( -\dfrac{4}{5} \right)+2\pi $ or $\Rightarrow x=\pi -{{\sin }^{-1}}\left( -\dfrac{4}{5} \right)$
Therefore, for the equation $\cos x-2\sin x=1$ , we get five required value of x which is equal to $x=0\text{, }\pi \text{, }2\pi \text{, }{{\sin }^{-1}}\left( -\dfrac{4}{5} \right)+2\pi \text{, }\pi -{{\sin }^{-1}}\left( -\dfrac{4}{5} \right)$ for the interval $0\le x\le 2\pi $ .

Note: In this question, we have used many trigonometric identities and formulas, therefore mention them in all the steps, whenever it is used. Do mention all the equation numbers properly and while explaining your step, mention the equation number to avoid errors while solving this problem.