Find all solutions of the equation $2 \sin ^{2} x-\cos x=1$ in the interval $[0,2 \pi)$.
The answer is $x_{1}=$ , $x_{2}=$ and $x_{3}=$ with $x_{1}< x_{2}< x_{3}$.
Answer
Final Answer: The solutions of the equation \(2 \sin ^{2} x-\cos x=1\) in the interval \([0,2 \pi)\) are \(x_{1}=\boxed{1.04719755119660}\), \(x_{2}=\boxed{3.14159265358979}\), and \(x_{3}=\boxed{5.23598775598299}\).
Step 1 ：Given the equation \(2 \sin ^{2} x-\cos x=1\).
Step 2 ：We know that \(\sin ^{2} x = 1 - \cos ^{2} x\). So we substitute this into the equation to get a quadratic equation in terms of \(\cos x\).
Step 3 ：The quadratic equation is \(-2\cos_x^2 - \cos_x + 1 = 0\).
Step 4 ：Solving this quadratic equation, we find the values of \(\cos x\) to be \(-1\) and \(1/2\).
Step 5 ：Using the inverse cosine function, we find the values of \(x\) in the interval \([0,2 \pi)\) to be approximately \(1.04719755119660\), \(3.14159265358979\), and \(5.23598775598299\).
Step 6 ：Sorting these values, we get \(x_{1}=1.04719755119660\), \(x_{2}=3.14159265358979\), and \(x_{3}=5.23598775598299\).
Step 7 ：Final Answer: The solutions of the equation \(2 \sin ^{2} x-\cos x=1\) in the interval \([0,2 \pi)\) are \(x_{1}=\boxed{1.04719755119660}\), \(x_{2}=\boxed{3.14159265358979}\), and \(x_{3}=\boxed{5.23598775598299}\).
