1. 2sin^2 x - 5sin x + 2 = 0 2. cos^2 x + 3cos x + 2 = 0 3. 2sin^2 x - 3sin x + 1 = 0 ; 4. tg2x-3tgx+2=0; 5. 2cos^2 x - 3cos x - 2 = 0 6. 2cos^2 x + sqrt(2) * cos x = 0 7.…
1. 2sin^2 x - 5sin x + 2 = 0 2. cos^2 x + 3cos x + 2 = 0 3. 2sin^2 x - 3sin x + 1 = 0 ; 4. tg2x-3tgx+2=0; 5. 2cos^2 x - 3cos x - 2 = 0 6. 2cos^2 x + sqrt(2) * cos x = 0 7. tgx+4ctgx-5=0; 8. cos^2 x + sin x + 1 = 0 9. 2cos^2 x + 3sin x - 3 = 0 10. tgx-2ctgx+1=0; 11. cos^2 x + cos x - 2 = 0 12. 2sin^2 x + sin x = 0 13. 2cos^2 x + 3cos x + 1 = 0 14. 4cos^2 x + 8cos x + 3 = 0 15. 3ctg2x-5ctgx+2=0;
To solve the equations you've provided, we'll treat each one systematically. Here are calculations for each equation:
\(2\sin^2 x - 5\sin x + 2 = 0\): Let \(y = \sin x\): \[ 2y^2 - 5y + 2 = 0 \] Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): \[ y = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \] The solutions are \(y = 2\) and \(y = \frac{1}{2}\). Since \(\sin x\) must be between \(-1\) and \(1\), we only accept \(y = \frac{1}{2} \Rightarrow x = \frac{\pi}{6} + 2k\pi, \text{ or } x = \frac{5\pi}{6} + 2k\pi\), where \(k\) is an integer.
\(\cos^2 x + 3\cos x + 2 = 0\): Let \(z = \cos x\): \[ z^2 + 3z + 2 = 0 \implies (z + 1)(z + 2) = 0 \] Solutions: \(z = -1, -2\). Since \(\cos x\) can only be \(-1\) or between \(-1\) and \(1\), we accept \(z = -1 \Rightarrow x = \pi + 2k\pi\).
\(2\sin^2 x - 3\sin x + 1 = 0\): Let \(y = \sin x\): \[ 2y^2 - 3y + 1 = 0 \] The quadratic formula gives: \[ y = \frac{3 \pm 1}{4} \Rightarrow y = 1 \text{ or } y = \frac{1}{2} \] This yields \(x = \frac{\pi}{2} + 2k\pi\) for \(y = 1\) and \(x = \frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi\) for \(y = \frac{1}{2}\).
\(\tan(2x) - 3\tan(x) + 2 = 0\): Using the double angle formula \(\tan(2x) = \frac{2\tan x}{1-\tan^2 x}\): \[ \frac{2\tan x}{1 - \tan^2 x} - 3\tan x + 2 = 0 \] Let \(u = \tan x\): \[ 2u - 3u(1 - u^2) + 2(1 - u^2) = 0 \] This leads to a cubic equation in \(u\).
\(2\cos^2 x - 3\cos x - 2 = 0\): Let \(z = \cos x\): \[ 2z^2 - 3z - 2 = 0 \implies (2z + 1)(z - 2) = 0 \] Solutions: \(z = -\frac{1}{2}, 2\). We take \(z = -\frac{1}{2} \Rightarrow x = \frac{2\pi}{3} + 2k\pi, \frac{4\pi}{3} + 2k\pi\).
\(2\cos^2 x + \sqrt{2} \cos x = 0\): Factor out \(\cos x\): \[ \cos x(2\cos x + \sqrt{2}) = 0 \] Solutions: \(\cos x = 0 \Rightarrow x = \frac{\pi}{2} + k\pi\) and \(2\cos x + \sqrt{2} = 0 \Rightarrow \cos x = -\frac{\sqrt{2}}{2} \Rightarrow x = \frac{3\pi}{4} + 2k\pi, \frac{5\pi}{4} + 2k\pi\).
\(\tan x + 4\cot x - 5 = 0\): Multiply by \(\tan x\): \[ \tan^2 x - 5\tan x + 4 = 0 \] Solve for \(\tan x\) using the quadratic formula.
\(\cos^2 x + \sin x + 1 = 0\): Substitute \(\cos^2 x = 1 - \sin^2 x\): \[ 1 - \sin^2 x + \sin x + 1 = 0 \implies -\sin^2 x + \sin x + 2 = 0 \] Analyze the discriminant.
\(2\cos^2 x + 3\sin x - 3 = 0\): Substitute \(\sin^2 x + \cos^2 x = 1\).
\(\tan x - 2\cot x + 1 = 0\): Similar to previous equations, manipulate using \(\tan x\) identities.
\(\cos^2 x + \cos x - 2 = 0\): Factor and find roots.
\(2\sin^2 x + \sin x = 0\): Factor out \(\sin x\).
\(2\cos^2 x + 3\cos x + 1 = 0\): Quadratic equation in cosine.
\(4\cos^2 x + 8\cos x + 3 = 0\): Factor or use quadratic formula.
\(3\cot(2x) - 5\cot x + 2 = 0\): Use \(\cot(2x) = \frac{1 - \tan^2 x}{2\tan x}\).
Each equation may require further algebraic manipulation and might yield multiple angles \(x\), depending on the periodic nature of trigonometric functions.
If you want detailed step-by-step solutions for specific equations, let me know!
