Question

How do you solve for \[x\] in \[4\sin x\cos x+2\sin x-2\cos x-1=0\]?

Answer 1

Hint: From the question given, we have been asked to solve for \[x\] in \[4\sin x\cos x+2\sin x-2\cos x-1=0\]. We can solve the above given trigonometric equation from the question by factoring the above given equation. After factoring the above given equation, then we have to solve the factors obtained to get the final solution for the given question.

Complete step-by-step answer:
First of all, from the question given, it has been given that \[4\sin x\cos x+2\sin x-2\cos x-1=0\]
We have to factor the above equation to get the solution for the given equation.
We can factor the given equation into \[\Rightarrow \left( 2\sin x-1 \right)\left( 2\cos x+1 \right)=0\]
Now, as we have already discussed earlier, we have to solve the above obtained two factors using standard trigonometric values.
Therefore \[\Rightarrow 2\sin x-1=0\]
Now, shift \[-1,2\] from the left hand side of the equation to the right hand side of the equation.
By shifting \[-1,2\] from left hand side of the equation to the right hand side of the equation, we get
\[\Rightarrow \sin x=\dfrac{1}{2}\]
\[\Rightarrow x={{30}^{0}}\text{ or x=15}{{\text{0}}^{0}}\]
And now \[\Rightarrow 2\cos x+1=0\]
Now, shift \[1,2\] from the left hand side of the equation to the right hand side of the equation.
By shifting \[1,2\] from left hand side of the equation to the right hand side of the equation, we get
\[\Rightarrow \cos x=-\dfrac{1}{2}\]
\[\Rightarrow x={{120}^{0}}\text{ or x=24}{{\text{0}}^{0}}\]
Therefore we can say that the value is given as \[\text{x}\in \left\{ {{30}^{0}},{{120}^{0}},{{150}^{0}},{{240}^{0}} \right\}\pm \text{any integral multiple of 3}{{60}^{\circ }}\].

Note: We should be very careful while factoring the given equation. Also, we should be very careful while solving the factors. Also, we should be very careful while doing the calculation part. Also, we should be well aware of the general trigonometric values. Also, we should be well known about the values of trigonometric angles. While deriving factors we can verify if the factors we derived are right or wrong by multiplying them and comparing with the equation. In this case if we verify we will have
\[\begin{align}
  & \left( 2\sin x-1 \right)\left( 2\cos x+1 \right)=0 \\
 & \Rightarrow 4\sin x\cos x+2\sin x-2\cos x-1=0 \\
\end{align}\]