[ 2(1 - \cos^2 x) - \cos x - 1 = 0 ] [ 2 - 2\cos^2 x - \cos x - 1 = 0 ] [ -2\cos^2 x - \cos x + 1 = 0 ] Multiply by -1: [ 2\cos^2 x + \cos x - 1 = 0 ]
Let ( t = \cos x ): [ 2t^2 + t - 1 = 0 \implies t = \frac-1 \pm \sqrt1 + 84 = \frac-1 \pm 34 ] [ t_1 = \frac24 = \frac12,\quad t_2 = \frac-44 = -1 ]
Use identity (\sin^2 x = 1 - \cos^2 x):
Introduction: Why Choose Langtang Valley Trek Nestled within Langtang National Park in Nepal’s Bagmati Province, just 80 km north of Kathmand...
From the moment I first saw the Himalayas, I knew my life would never be the same. It wasn’t just a landscape; it was a call, a silent invitation to e... [ 2(1 - \cos^2 x) - \cos x
The Langtang Valley Trek in Nepal is often described as the perfect mix of adventure and cultural immersion. Nestled just north of Kathmandu, this tre... [ 2(1 - \cos^2 x) - \cos x
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[ 2(1 - \cos^2 x) - \cos x - 1 = 0 ] [ 2 - 2\cos^2 x - \cos x - 1 = 0 ] [ -2\cos^2 x - \cos x + 1 = 0 ] Multiply by -1: [ 2\cos^2 x + \cos x - 1 = 0 ]
Let ( t = \cos x ): [ 2t^2 + t - 1 = 0 \implies t = \frac-1 \pm \sqrt1 + 84 = \frac-1 \pm 34 ] [ t_1 = \frac24 = \frac12,\quad t_2 = \frac-44 = -1 ]
Use identity (\sin^2 x = 1 - \cos^2 x):
