Find all solutions of the equation in the interval [0,2pi ) -4cosx=-sin^2x+1 Write your answer in radians in terms of pi If there is more than one solution separate them with commas. x=[ ]

To solve the equation \(-4\cos x = -\sin^2 x + 1\) in the interval \([0, 2\pi)\), we can follow these steps:<br /><br />1. Rewrite the equation:<br /> \[<br /> -4\cos x = -\sin^2 x + 1<br /> \]<br /> \[<br /> 4\cos x = \sin^2 x - 1<br /> \]<br /> \[<br /> \cos x = \frac{\sin^2 x - 1}{4}<br /> \]<br /><br />2. Use the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\) to express \(\cos x\) in terms of \(\sin x\):<br /> \[<br /> \cos x = \sqrt{1 - \sin^2 x}<br /> \]<br /><br />3. Substitute \(\cos x\) into the equation:<br /> \[<br /> \sqrt{1 - \sin^2 x} = \frac{\sin^2 x - 1}{4}<br /> \]<br /><br />4. Square both sides to eliminate the square root:<br /> \[<br /> 1 - \sin^2 x = \left(\frac{\sin^2 x - 1}{4}\right)^2<br /> \]<br /> \[<br /> 1 - \sin^2 x = \frac{(\sin^2 x - 1)^2}{16}<br /> \]<br /><br />5. Multiply both sides by 16 to clear the fraction:<br /> \[<br /> 16(1 - \sin^2 x) = (\sin^2 x - 1)^2<br /> \]<br /> \[<br /> 16 - 16\sin^2 x = (\sin^2 x - 1)^2<br /> \]<br /><br />6. Expand the right-hand side:<br /> \[<br /> 16 - 16\sin^2 x = \sin^4 x - 2\sin^2 x + 1<br /> \]<br /><br />7. Rearrange the equation to form a quadratic in \(\sin^2 x\):<br /> \[<br /> \sin^4 x - 18\sin^2 x + 15 = 0<br /> \]<br /><br />8. Let \(u = \sin^2 x\), then the equation becomes:<br /> \[<br /> u^2 - 18u + 15 = 0<br /> \]<br /><br />9. Solve the quadratic equation using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):<br /> \[<br /> u = \frac{18 \pm \sqrt{324 - 60}}{2}<br /> \]<br /> \[<br /> u = \frac{18 \pm \sqrt{264}}{2}<br /> \]<br /> \[<br /> u = \frac{18 \pm 2\sqrt{66}}{2}<br /> \]<br /> \[<br /> u = 9 \pm \sqrt{66}<br /> \]<br /><br />10. Since \(\sin^2 x\) must be in the interval \([0, 1]\), we need to check which solutions are valid:<br /> \[<br /> u = 9 - \sqrt{66} \quad \text{and} \quad u = 9 + \sqrt{66}<br /> \]<br /> Only \(u = 9 - \sqrt{66}\) is in the interval \([0, 1]\).<br /><br />11. Find \(\sin x\) from \(u = \sin^2 x\):<br /> \[<br /> \sin^2 x = 9 - \sqrt{66}<br /> \]<br /> \[<br /> \sin x = \pm \sqrt{9 - \sqrt{66}}<br /> \]<br /><br />12. Determine the corresponding \(x\) values in the interval \([0, 2\pi)\):<br /> \[<br /> x = \arcsin(\sqrt{9 - \sqrt{66}}) \quad \text{and} \quad x = \pi - \arcsin(\sqrt{9 - \sqrt{66}})<br /> \]<br /> \[<br /> x = \pi + \arcsin(\sqrt{9 - \sqrt{66}}) \quad \text{and} \quad x = 2\pi - \arcsin(\sqrt{9 - \sqrt{66}})<br /> \]<br /><br />Thus, the solutions in radians in terms of \(\pi\) are:<br />\[<br />x = \arcsin(\sqrt{9 - \sqrt{66}}), \pi - \arcsin(\sqrt{9 - \sqrt{66}}), \pi + \arcsin(\sqrt{9 - \sqrt{66}}), 2\pi - \arcsin(\sqrt{9 - \sqrt{66}})<br />\]