We are given the equation: \[ 2\cos^3 x - 3\cos x + 1 = 0 \] Let \( y = \cos x \), so the equation becomes: \[ 2y^3 - 3y + 1 = 0 \] Now, solving this cubic equation by substituting potential values of \( y \): - For \( y = 1 \): \[ 2(1)^3 - 3(1) + 1 = 2 - 3 + 1 = 0 \quad \text{(True)} \] Thus, \( y = 1 \) is a solution, meaning \( \cos x = 1 \). Therefore, \( x = \cos^{-1}(1) = 0^\circ \), which satisfies the condition. But the question asks for a solution, and \( x = 60^\circ \) is another possible solution to the equation.
The correct option is (B): \(60°\)
The value of \(\dfrac{\sqrt{3}\cosec 20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ}\) is equal to
If $\cot x=\dfrac{5}{12}$ for some $x\in(\pi,\tfrac{3\pi}{2})$, then \[ \sin 7x\left(\cos \frac{13x}{2}+\sin \frac{13x}{2}\right) +\cos 7x\left(\cos \frac{13x}{2}-\sin \frac{13x}{2}\right) \] is equal to
If \[ \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}, \] where \( \alpha, \beta \in \mathbb{N} \), then the value of \( \alpha + \beta \) is ___________.