cos42θ+2sin22θ=17(cosθ+sinθ)8 if θ = - Infinity Learn by Sri Chaitanya

A
$105 °$
B
$175 °$
C
$280 °$
D
$340 °$

Solution:

${(1 - \sin^{2} 2 θ)}^{2} + 2 \sin^{2} 2 θ = 17 (1 + \sin 2 θ)^{4}$
$\Rightarrow 1 + x^{4} = 17 (1 + x)^{4},$ where $x = \sin 2 θ$

or $x^{2} + \frac{1}{x^{2}} = 17 {(x + \frac{1}{x} + 2)}^{2}$

If $t = x + \frac{1}{x},$ we get $t^{2} - 2 = 17 (t + 2)^{2}$
$\Rightarrow 8 t^{2} + 34 t + 35 = 0 ∴ x + \frac{1}{x} = t = - \frac{5}{2}, - \frac{7}{4}$
But $x + \frac{1}{x}$ cannot lie between -2 and 2,

$\begin{aligned} x + \frac{1}{x} = - \frac{5}{2} \Rightarrow x = \sin 2 θ = - \frac{1}{2} \\ ∴ θ = 105^{\circ}, 165^{\circ}, 285^{\circ}, 345^{\circ} \end{aligned}$

$\cos^{4} 2 θ + 2 \sin^{2} 2 θ = 17 (\cos θ + \sin θ)^{8}$ if $θ =$

Solution:

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