What is the general solution for 14sin(x+pi/2)+21tan(pi-x)=0 ?

The general solution for 14sin(x+pi/2)+21tan(pi-x)=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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14sin(x+pi/2)+21tan(pi-x)=0

14 sin (x + \frac{π}{2}) + 21 tan (π - x) = 0

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solution steps

14 sin (x + \frac{π}{2}) + 21 tan (π - x) = 0

Rewrite using trig identities

14 cos (x) - \frac{21 sin ( x )}{cos ( x )} = 0

Simplify

14 cos (x) - \frac{21 sin ( x )}{cos ( x )} : \frac{14 cos ^{2} ( x ) - 21 sin ( x )}{cos ( x )}

\frac{14 cos ^{2} ( x ) - 21 sin ( x )}{cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

14 cos^{2} (x) - 21 sin (x) = 0

Rewrite using trig identities

(1 - sin^{2} (x)) \cdot 14 - 21 sin (x) = 0

Solve by substitution

sin (x) = - 2, sin (x) = \frac{1}{2}

sin (x) = - 2 :

No Solution

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

tan (x) = - \frac{15}{8}

8 cos^{3} (x) = 8 cos (x)

4 cos (2 θ) = cos^{2} (θ) - 2

2 cos^{2} (x) + sin (x) = 1, 0 \leq x \leq 2 π

cos (x) = \frac{5}{5}

What is the general solution for 14sin(x+pi/2)+21tan(pi-x)=0 ?
The general solution for 14sin(x+pi/2)+21tan(pi-x)=0 is x= pi/6+2pin,x=(5pi)/6+2pin