The equation 4Sin2x + 4Sinx + a2 – 3 = 0 has a solution if

The equation 4Sin²x + 4Sinx + a² - 3 = 0 has a solution if

A
$- 2 \leq a \leq 2$
B
$- 1 \leq a \leq 1$
C
$- 3 \leq a \leq 3$
D
$- 4 \leq a \leq 4$

Solution:

$\sin x = \frac{- 4 \pm \sqrt{16 - 16 (a^{2} - 3)}}{8}$

$- 1 \leq \frac{- 1 \pm \sqrt{4 - a^{2}}}{2} \leq 1$

$- 2 \leq - 1 \pm \sqrt{4 - a^{2}} \leq 2$

$0 \leq \sqrt{4 - a^{2}} \leq 3$

$4 - a^{2} \geq 0 \Rightarrow a \in [- 2, 2]$

Related content

Test your English Vocabulary

CUET Exam Dates 2024 – Application Form, Fees, Eligibility

CBSE Class 12 IP Answer Key 2024,Informatics Practices Paper Solution For SET 1, 2, 3, 4

CUET UG Cut Off 2024, Category, Universities and Colleges Wise Expected Cut Off

Modal Verbs

Helping Verbs

Letter To Your Friend About Your School Trip

Action Verbs

CUET 2024 – List of Colleges and Participating Universities Accepting CUET Exam Score

SRMJEEE Online Test Series – Practice Papers