Question : If $\theta$ is positive acute angle and $7\cos^{2}\theta+3\sin^{2}\theta =4$, then the value of $\theta$ is:
Option 1: $60^{\circ}$
Option 2: $30^{\circ}$
Option 3: $45^{\circ}$
Option 4: $90^{\circ}$
Correct Answer: $60^{\circ}$
Solution :
$7\cos^{2}\theta +3\sin^{2}\theta = 4$
$⇒7\cos^{2}\theta +3(1-\cos^{2}\theta) = 4$ ........[we know that $\sin^{2}\theta + \cos^{2}\theta=1$]
$⇒4\cos^{2}\theta +3 = 4$
$⇒\cos^{2}\theta = \frac{1}{4}$
$⇒\cos^{2}\theta - \frac{1}{4}$ = 0
$⇒(\cos\theta + \frac{1}{2})(\cos\theta - \frac{1}{2})$ = 0
$⇒\cos\theta =\frac{1}{2}\ \text{or,} -\frac{1}{2}$
$⇒\cos\theta =\cos 60^{\circ}\ \text{or,}\cos 120^{\circ}$
Since $\theta$ is a positive acute angle,
So, $\theta = 60^{\circ}$
Hence, the correct answer is $60^{\circ}$.
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