Question

In the equation 6${sin}^2$$\theta$ - 11sin$\theta$ + 4 = 0, show that one value of $\theta$ is absurd and find the other value.

• A. 30^o
• B. 60^o
• C. 45^o
• D. None of these

Answer 1

The correct option is A

30^o

$6si{n}^2\theta -11sin\theta +4$ = 0

$6si{n}^2\theta -8sin\theta -3sin\theta +4$ = 0

i.e. $2sin\theta (3sin\theta -4)-1(3sin\theta -4)$ = 0

i.e. $(2sin\theta -1)(3sin\theta -4)$ = 0

$\therefore$$\theta$ = ${30}^{\circ }$ or $sin\theta$ = $\frac{4}{3}$ is inadmissible as the value of sine of any angle cannot be numerically greater than 1.