Answer
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Hint: In this problem, we have to find all the solutions within the given interval for the given trigonometric equation. We can first take the left-hand side and change the given terms with a trigonometric formula, then we can use a quadratic formula to find the value of x.
Complete step by step solution:
We know that the given trigonometric expression is,
\[4{{\sin }^{2}}x=2\cos x+1\]
We can convert the left-hand side with the trigonometric formula,
We can write \[{{\sin }^{2}}x=1-{{\cos }^{2}}x\] , we get
\[\Rightarrow 4\left( 1-{{\cos }^{2}}x \right)=2\cos x+1\]
We can now multiply the number inside the brackets, we get
\[\Rightarrow \left( 4-4{{\cos }^{2}}x \right)=2\cos x+1\]
We can now take the numbers form the left-hand side to the right-hand side by changing the sign and simplify it, we get
\[\Rightarrow 4{{\cos }^{2}}x+2\cos x-3=0\] ….. (1)
Now we can use the quadratic formula.
We know that the quadratic formula for the equation is \[a{{x}^{2}}+bx+c=0\],
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
We can now compare the equation (1) and the general equation, we can get
a = 4, b = 2, c = -3
we can substitute the values in the quadratic formula we get
\[\Rightarrow \cos x=\dfrac{-2\pm \sqrt{{{2}^{2}}+4.4.3}}{2.4}\]
We can now simplify the above step, we get
\[\begin{align}
& \Rightarrow \cos x=\dfrac{-1\pm \sqrt{13}}{4} \\
& \Rightarrow x={{\cos }^{-1}}\left( \dfrac{-1\pm \sqrt{13}}{4} \right) \\
\end{align}\]
We know that cos x cannot be less than -1, so we can neglect the value \[x={{\cos }^{-1}}\left( \dfrac{-\sqrt{13}-1}{4} \right)\]
We have already given an interval \[\left[ 0,2\pi \right)\].
Therefore, the value of x within the given interval \[\left[ 0,2\pi \right)\] is
\[x={{\cos }^{-1}}\left( \dfrac{\sqrt{13}-1}{4} \right)\] and \[x=2\pi -{{\cos }^{-1}}\left( \dfrac{\sqrt{13}-1}{4} \right)\]
Note: Students make mistakes while solving using the quadratic equation, which should be concentrated. We should know some trigonometric rules and formulas to solve these types of problems. We should also know the formula for the quadratic equation to solve these types of problems.
Complete step by step solution:
We know that the given trigonometric expression is,
\[4{{\sin }^{2}}x=2\cos x+1\]
We can convert the left-hand side with the trigonometric formula,
We can write \[{{\sin }^{2}}x=1-{{\cos }^{2}}x\] , we get
\[\Rightarrow 4\left( 1-{{\cos }^{2}}x \right)=2\cos x+1\]
We can now multiply the number inside the brackets, we get
\[\Rightarrow \left( 4-4{{\cos }^{2}}x \right)=2\cos x+1\]
We can now take the numbers form the left-hand side to the right-hand side by changing the sign and simplify it, we get
\[\Rightarrow 4{{\cos }^{2}}x+2\cos x-3=0\] ….. (1)
Now we can use the quadratic formula.
We know that the quadratic formula for the equation is \[a{{x}^{2}}+bx+c=0\],
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
We can now compare the equation (1) and the general equation, we can get
a = 4, b = 2, c = -3
we can substitute the values in the quadratic formula we get
\[\Rightarrow \cos x=\dfrac{-2\pm \sqrt{{{2}^{2}}+4.4.3}}{2.4}\]
We can now simplify the above step, we get
\[\begin{align}
& \Rightarrow \cos x=\dfrac{-1\pm \sqrt{13}}{4} \\
& \Rightarrow x={{\cos }^{-1}}\left( \dfrac{-1\pm \sqrt{13}}{4} \right) \\
\end{align}\]
We know that cos x cannot be less than -1, so we can neglect the value \[x={{\cos }^{-1}}\left( \dfrac{-\sqrt{13}-1}{4} \right)\]
We have already given an interval \[\left[ 0,2\pi \right)\].
Therefore, the value of x within the given interval \[\left[ 0,2\pi \right)\] is
\[x={{\cos }^{-1}}\left( \dfrac{\sqrt{13}-1}{4} \right)\] and \[x=2\pi -{{\cos }^{-1}}\left( \dfrac{\sqrt{13}-1}{4} \right)\]
Note: Students make mistakes while solving using the quadratic equation, which should be concentrated. We should know some trigonometric rules and formulas to solve these types of problems. We should also know the formula for the quadratic equation to solve these types of problems.
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