Number of solutions of the equation -sin4-1 in -0-2- is

|sin4θ|=1 ⇒sin4θ=±1 When sin4θ=1 ⇒4θ=(4n+1)π2, n∈ℤ ⇒θ=(4n+1)π8, n∈ℤ Number of solutions in [0,2π] are 4. i.e. {π8,5π8,9π8,13π8} When sin4θ= 1 ⇒4θ=(4n 1)π2, n∈ ...

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Standard XII

Mathematics

Principal Solution of Trigonometric Equation

Number of sol...

Question

Number of solutions of the equation $| sin 4 θ | = 1$ in $θ \in [0, 2 π]$ is

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Solution

$| sin 4 θ | = 1$
$\Rightarrow sin 4 θ = \pm 1$

When $sin 4 θ = 1$
$\Rightarrow 4 θ = (4 n + 1) \frac{π}{2}, n \in Z$
$\Rightarrow θ = (4 n + 1) \frac{π}{8}, n \in Z$
Number of solutions in $[0, 2 π]$ are $4.$
i.e. ${\frac{π}{8}, \frac{5 π}{8}, \frac{9 π}{8}, \frac{13 π}{8}}$

When $sin 4 θ = - 1$
$\Rightarrow 4 θ = (4 n - 1) \frac{π}{2}, n \in Z$
$\Rightarrow θ = (4 n - 1) \frac{π}{8}, n \in Z$
Number of solutions in $[0, 2 π]$ are $4.$
i.e. ${\frac{3 π}{8}, \frac{7 π}{8}, \frac{11 π}{8}, \frac{15 π}{8}}$

Hence, total number of solutions $= 8$

Alternate Solution:
We know that $| sin θ |$ is periodic with period $π .$
So, $| sin 4 θ |$ will be periodic with period $\frac{π}{4} .$
$∴$ Graph of $y = | sin 4 θ |$ in $θ \in [0, π]$ is

So, number of solution in $[0, π / 4]$ is $1.$
$∴$ Number of solution in $[0, 2 π]$ is $8.$

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Principal Solution of Trigonometric Equation

Standard XII Mathematics

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Number of solutions of the equation |sin4θ|=1 in θ∈[0,2π] is

Number of solutions of the equation $| sin 4 θ | = 1$ in $θ \in [0, 2 π]$ is