- Home
- Standard 11
- Mathematics
Trigonometrical Equations
normal
The real roots of the equation $cos^7x\, +\, sin^4x\, =\, 1$ in the interval $(-\pi, \pi)$ are
A
$ \{- \frac{\pi }{2}\,,\,0 \}$
B
$\{ - \frac{\pi }{2}\,,\,0\,,\,\frac{\pi }{2} \}$
C
$\{ \frac{\pi }{2}\,,\,0 \}$
D
$\{ 0\,\,,\,\,\frac{\pi }{4}\,\,,\,\frac{\pi }{2} \}$
Solution
$\cos ^{7} x+\sin ^{4} x=1$
$\cos ^{7} x=1-\sin ^{4} x=\left(1-\sin ^{2} x\right)\left(1+\sin ^{2} x\right)$
$\cos ^{7} x=\cos ^{2} x\left(1+\sin ^{2} x\right)$
$\cos ^{7}(x)-\cos ^{2} x\left(2-\cos ^{2} x\right)=0$
$\cos ^{2} x\left(\cos ^{5} x+\cos ^{2}(x)-2\right)=0$
$\cos ^{2}(x)=0$ implies
$x=\frac{\pm \pi}{2}$
And
$\cos ^{5} x+\cos ^{2} x-2=0$ implies
$\cos (x)=1$
$x=0$
Hence
$x \in\left\{\frac{-\pi}{2}, 0, \frac{\pi}{2}\right\}$
Standard 11
Mathematics
