MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Domain

Solve |sin ...

Question

Solve $| sin x + cos x | = | sin x | + | cos x |, x \in [0, 2 π]$ .

Open in App

Solution

Given $| s i n x + c o s x | = | s i n x | + | c o s x |$
For $0 \leq x \leq π / 2$
$s i n x + c o s x = s i n x + c o s x$
$\Rightarrow 0 ⩽ x ⩽ π / 2$
For $π / 2 ⩽ x ⩽ π, s i n x > 0 c o s x < 0$
LHS $s i n x - c o s x > 0 x < 3 π / 4$
$s i n x - c o s x < 0 x > 3 π / 2$
RHS : $s i n x + c o s x$
L.H.S $\neq$ RHS
For $π ⩽ x < 3 π / 2$
LHS = $| - s i n x - c o s x |$
RHS = $| - s i n x | + | - c o s x |$
LHS = RHS $\Rightarrow π ⩽ x ⩽ 3 π / 2$
For $3 π / 2 < x < 2 π$
LHS : $| - s i n x + c o s x | = | c o s x - s i n x |$
RHS : $| s i n x | + | c o s x |$
LHS $\neq$ RHS
$\Rightarrow 0 ⩽ x ⩽ π / 2$ and $π ⩽ x ⩽ 3 π / 2$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Solve the following equations:
(i)

\sin x + \cos x = \sqrt{2}

\sqrt{3} \cos x + \sin x = 1

\sin x + \cos x = 1

cosec x = 1 + \cot x

f (x) = sin x + | cos x | - ∣ ∣ s i n x - | c o s x | ∣ ∣

x ϵ (0, 2 π)

then the number of points of non-differentiability of

f (x)

Q. Solve for x if
\(

∣ ∣ ∣ ∣ \begin{matrix} s i n x & c o s x & c o s x c o s x & s i n x & c o s x c o s x & c o s x & s i n x \end{matrix} ∣ ∣ ∣ ∣

\int \frac{sin x + cos x}{\sqrt{1 + sin 2 x}} d x

, Given that x takes values for

sin x + cos x \geq 0

t a n^{- 1} {\frac{c o s x + s i n x}{c o s x - s i n x}}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon