SATHEE: Trigonometrical Equations 1 Question 1

Trigonometrical Equations 1 Question 1

1. Let $S$ be the set of all $α \in R$ such that the equation, $\cos 2 x + α \sin x = 2 α - 7$ has a solution. Then, $S$ is equal to

(2019 Main, 12 April II)

(a) $R$

(b) $[1, 4]$

(d) $[2, 6]$

Show Answer

Answer:

Correct Answer: 1. (d)

Solution:

The given trigonometric equation is

$\begin{array}{lc} \cos 2 x + α \sin x = 2 α - 7 \\ \Rightarrow & 1 - 2 \sin^{2} x + α \sin x = 2 α - 7 \\ [∵ \cos 2 x = 1 - 2 \sin^{2} x] \\ \Rightarrow & 2 \sin^{2} x - α \sin x + 2 α - 8 = 0 \\ \Rightarrow & 2 (\sin^{2} x - 4) - α (\sin x - 2) = 0 \\ \Rightarrow & 2 (\sin x - 2) (\sin x + 2) - α (\sin x - 2) = 0 \end{array}$

$\begin{aligned} \Rightarrow (\sin x - 2) (2 \sin x + 4 - α) = 0 \\ ∴ 2 \sin x + 4 - α = 0 \\ \Rightarrow \sin x = \frac{α - 4}{2} \end{aligned}$

$[∵ \sin x - 2 \neq 0]$

Now, as we know $- 1 \leq \sin x \leq 1$

$\begin{array}{ll} ∴ & - 1 \leq \frac{α - 4}{2} \leq 1 \\ \Rightarrow & - 2 \leq α - 4 \leq 2 \Rightarrow 2 \leq α \leq 6 \Rightarrow α \in [2, 6] \end{array}$

Trigonometrical Equations 1 Question 1

1. Let $S$ be the set of all $α \in R$ such that the equation, $\cos 2 x + α \sin x = 2 α - 7$ has a solution. Then, $S$ is equal to

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Trigonometrical Equations 1 Question 1

1. Let S be the set of all α∈R such that the equation, cos⁡2x+αsin⁡x=2α−7 has a solution. Then, S is equal to

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

1. Let $S$ be the set of all $α \in R$ such that the equation, $\cos 2 x + α \sin x = 2 α - 7$ has a solution. Then, $S$ is equal to