Question : The value of $\theta$ $ \left ( 0\leq \theta \leq 90^{\circ} \right )$ satisfying $2\sin^{2}\theta = 3\cos \theta$ is:
Option 1: $60^{\circ}$
Option 2: $30^{\circ}$
Option 3: $90^{\circ}$
Option 4: $45^{\circ}$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $60^{\circ}$
Solution : Given that, $2\sin^{2}\theta = 3\cos\theta$. Use the identity $\sin^{2}\theta + \cos^{2}\theta = 1$ $⇒2(1-\cos^{2}\theta) = 3\cos\theta$ $⇒2\cos^{2}\theta + 3\cos\theta - 2 = 0$ $⇒(2\cos\theta - 1)(\cos\theta + 2) = 0$ $⇒\cos\theta = \frac{1}{2}$ or $\cos\theta = -2$ Since $0^{\circ} \leq \theta \leq 90^{\circ}$, we can take $\cos\theta = \frac{1}{2}$. $⇒\cos\theta = \frac{1}{2}=\cos60^{\circ}$ $\Rightarrow \theta = 60^{\circ}$ Hence, the correct answer is $60^{\circ}$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $0\leq\theta\leq 90^{\circ}$ and $4\cos^{2}\theta-4\sqrt{3}\cos\theta+3=0$, then the value of $\theta$ is:
Question : If $\sin ^2 \theta-3 \sin \theta+2=0$, then find the value of $\theta\left(0^{\circ} \leq \theta \leq 90^{\circ}\right)$.
Question : If $\left(\frac{\cos A}{1-\sin A}\right)+\left(\frac{\cos A}{1+\sin A}\right)=4$, then what will be the value of $A$? $\left(0^{\circ}<\theta<90^{\circ}\right)$
Question : The value of $\left(\sin 30^{\circ} \cos 60^{\circ}-\cos 30^{\circ} \sin 60^{\circ}\right)$ is equal to:
Question : If $\sin\left ( 2a+45^{\circ} \right )=\cos\left ( 30^{\circ}-a \right )$ where $0^{\circ}< a< 90^{\circ}$, then the value of a is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile