The correct answer is: 32
\(3 \cos^2(2\theta) + 6 \cos^2(\theta) - 10\cos(2\theta) + 5 = 0\)\(-\frac{10(1+cos2θ)}{2}\)
\(3\cos^2(2\theta) + \cos^2(\theta) = 0\)
\(\cos^2(\theta) = 0 \quad \text{or} \quad \cos^2(\theta) =\) \(-\frac{1}{3}\)
As \(\theta \in [0, \pi], \quad \cos(2\theta) = -\frac{1}{3} ⇒ 2 times\)
\(⇒\) \(\theta \in [-4\pi, 4\pi], \quad \cos(2\theta) = -\frac{1}{3}\)\(-\frac{1}{3}\) \(⇒\) 16 times
Similarly, \(\cos(2\theta) = 0\) \(⇒ 16\) times
∴ Total is 32 solutions.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank is _________ cm. (Take \( g = 10 \, {m/s}^2 \)).
The relationship between the sides and angles of a right-angle triangle is described by trigonometry functions, sometimes known as circular functions. These trigonometric functions derive the relationship between the angles and sides of a triangle. In trigonometry, there are three primary functions of sine (sin), cosine (cos), tangent (tan). The other three main functions can be derived from the primary functions as cotangent (cot), secant (sec), and cosecant (cosec).
sin x = a/h
cos x = b/h
tan x = a/b
Tan x can also be represented as sin x/cos x
sec x = 1/cosx = h/b
cosec x = 1/sinx = h/a
cot x = 1/tan x = b/a